Can you measure height with a stopwatch? How could that be possible?
This lesson is all about quadratic functions, like the one used in Physics and described later in the lesson, which allows a good approximation of the height of a tall object to be calculated with time!
Watch This
This lesson focuses on the use of multiple methods for solving quadratic equations. Of the methods discussed, the "Factoring to Solve" and "Completing the Square" methods commonly require more practice and explanation.
You may choose to watch both videos below, or only the one referring to the solution method you are less comfortable with.
SOLVE BY FACTORING:
James Sousa: Solving Quadratic Equations by Factoring
SOLVE BY COMPLETING THE SQUARE:
James Sousa: Completing the Square to Solve Quadratic Equations
Guidance
A quadratic function can be described as:

Formally: A function \begin{align*}f\end{align*}
f defined by \begin{align*}f(x)=ax^{2}+bx+c\end{align*}f(x)=ax2+bx+c , where \begin{align*}a, b,\end{align*}a,b, and \begin{align*}c\end{align*}c are real numbers and \begin{align*}a \ne 0\end{align*}a≠0 .  Informally: The defining characteristic of a quadratic function is that it is a polynomial whose highest exponent is 2.
There are several ways to write quadratic functions:

standard form, the form of the quadratic function above: \begin{align*}f(x)=ax^{2}+bx+c\end{align*}
f(x)=ax2+bx+c 
vertex form, commonly used for quick sketching: \begin{align*}f(x)=a(xh)^{2}+k\end{align*}
f(x)=a(x−h)2+k 
factored form, excellent for finding xintercepts: \begin{align*}f(x)=a(xr_{1})(xr_{2})\end{align*}
f(x)=a(x−r1)(x−r2)
You can convert between forms of quadratic functions using algebra and you will see that there are uses for each of these forms when working with quadratic functions.
When graphing, the \begin{align*}y\end{align*}
Example A
Solve \begin{align*}x^{2}5x+6=0\end{align*}
Solution
The factor method is based on writing the quadratic equation in factored form. That is, as a product of two linear expressions. So our equation maybe solved by the following way:
\begin{align*}x^{2}5x+6 & = 0\\ (x3)(x2) & = 0\end{align*}
Recall, that
\begin{align*}a \cdot b=0 \ \text{if and only if} \ a=0 \ \text{and} \ b=0\end{align*}
This tells us that
\begin{align*}x3=0\end{align*}
or
\begin{align*}x2=0\end{align*}
which give the roots (or zeros)
\begin{align*}x=3\end{align*}
and
\begin{align*}x=2\end{align*}
In other words, the solution set is {2, 3}.
Example B
Solve \begin{align*}x^{2}5x+6=0\end{align*}
Solution
To solve the above equation by completing the square, first move the “\begin{align*}c\end{align*}
\begin{align*}x^{2}5x=6\end{align*}
Next, make the lefthand side a “perfect square” by adding the appropriate number. To do so, take onehalf of the coefficient of \begin{align*}x\end{align*}
\begin{align*}b & = 5 \\ \frac{b}{2} & = \frac{5}{2}\\ \left ( \frac{b}{2} \right ) ^2 & = \frac{25}{4}\end{align*}
Adding to both sides of the equation,
\begin{align*}x^{2}5x+\frac{25}{4} & = 6 + \frac{25}{4}\\ x^2 5x + \frac{25}{4} & = \frac{1}{4}\\ \left ( x\frac{5}{2} \right )^2 & = \frac{1}{4}\end{align*}
this last equation can be easily solved by taking the square root of both sides,
\begin{align*}\left(x\frac{5}{2}\right)=\sqrt{\frac{1}{4}}=\pm\frac{1}{2}\end{align*}
Hence
\begin{align*}x=\pm\frac{1}{2}+\frac{5}{2}\end{align*}
and the solutions are
\begin{align*}x=3\end{align*}
and
\begin{align*}x=2\end{align*}
which are identical to answers of the factor method.
Example C
Solve \begin{align*}x^{2}5x+6=0\end{align*}
Solution
The quadratic formula works for finding the xintercepts in all quadratic equations, therefore it is highly encouraged that you memorize the formula. The other methods can be much faster though, so it is well worth understanding each of the different methods.
Recall, if \begin{align*}ax^{2}+bx+c=0\end{align*}
\begin{align*}x=\frac{b\pm\sqrt{b^{2}4ac}}{2a}.\end{align*}
Here the coefficients are \begin{align*}a=1, b=5,\end{align*}
Substituting into the quadratic formula, we get:
\begin{align*}x & = \frac{(5) \pm \sqrt{(5)^2  4(1)(6)}}{2(1)}\\ & = \frac{5 \pm \sqrt{2524}}{2}\\ & = \frac{5 \pm 1}{2}\\ & = 3 \ \text{or} \ 2\end{align*}
which is, again, identical to our two solutions above.
Were you able to solve the problem at the beginning of the lesson? What formula could be used to calculate the height of an object based on the time required for an object to fall from the top?
This is not completely accurate, as it ignores the effects of air resistance of an object, but offers a very close approximate for most common objects. The value g is gravity. In Metric units this is \begin{align*}9.81 m/s/s\end{align*} The value t is the time taken for the object to fall in seconds The result h is the distance or the height in meters 

Vocabulary
Completing the Square: A common method for solving quadratic equations accomplished by making a perfect square trinomial on one side of the equation.
Factor to Solve: A common method for solving quadratic equations accomplished by factoring a trinomial into two binomials and identifying the values of x that make each binomial equal to zero.
Quadratic Formula The most reliable method of solving for x intercepts in a quadratic equation, though not always the most efficient, involves substituting the coefficients of each of the terms of the quadratic function into the formula: \begin{align*}x=\frac{b\pm\sqrt{b^{2}4ac}}{2a}.\end{align*}
Guided Practice
Questions
1) Solve each equation by factoring.
 a. \begin{align*}x^{2}=64\end{align*}
 b. \begin{align*}4x^{2}+7x=2\end{align*}
 c. \begin{align*}4x^{2}17x=4\end{align*}
2) Solve each equation by completing the square
 a. \begin{align*}x^{2}4x+1=0\end{align*}
 b. \begin{align*}\frac{2}{3}x^{2}x+\frac{1}{3}=0\end{align*}
 c. \begin{align*}x^{2}+2.8=4.7x\end{align*}
3) Solve each equation by using the quadratic formula
 a. \begin{align*}0.4x^{2}+x0.3=0\end{align*}
 b. \begin{align*}25x^{2}+80x+61=0\end{align*}
 c. \begin{align*}(z+6)^{2}+2z=0\end{align*}
Answers
1) Factoring to Solve:
 a. \begin{align*}x = \sqrt{64} \therefore x = \pm8\end{align*}

b. \begin{align*}4x^{2} + 7x  2 = 0\end{align*}
 \begin{align*}0 = (4x  1)(x + 2) \therefore x = \frac{1}{4}, 2\end{align*}

c. \begin{align*}4x^{2}  17x + 4 = 0\end{align*}
 \begin{align*}0 = (4x  1)(x  4) \therefore x = \frac{1}{4}, 4\end{align*}
2) Completing the Square:

a. \begin{align*}x^{2}4x+4=3\end{align*}
 \begin{align*}(x2)(x2)=3\end{align*}
 \begin{align*}x = 2\pm\sqrt{3}\end{align*}

b. \begin{align*}x^{2}\frac{3}{2}x+\frac{1}{2}=0\end{align*}
 \begin{align*}x^{2}\frac{3}{2}x+\frac{9}{16}=\frac{9}{16}\frac{1}{2}\end{align*}
 \begin{align*}(x\frac{3}{4})(x\frac{3}{4}) = \frac{1}{16}\end{align*}
 \begin{align*}x=\frac{1}{2}, 1\end{align*}

c. \begin{align*}x^{2}  4.7x + 2.8 = 0\end{align*}
 \begin{align*}x^2  4.7x + 2.35^{2} = 2.35^{2}2.8\end{align*}
 \begin{align*}(x2.35)(x2.35)=2.7225\end{align*}
 \begin{align*}x2.35=\pm1.65\end{align*}
 \begin{align*}x=\frac{7}{10}, 4\end{align*}
3) Quadratic Formula:

a. \begin{align*}x = \frac{1\pm\sqrt{1^{2}4 \cdot 0.4 \cdot (0.3)}}{2\cdot0.4}\end{align*}
 \begin{align*}x=0.275, 2.78\end{align*}

b. \begin{align*}x = \frac{80\pm\sqrt{80^{2}4\cdot25\cdot61}}{2\cdot25}\end{align*}
 \begin{align*}x = \frac{8}{5}\pm\frac{\sqrt{3}}{5}\end{align*}

c. \begin{align*}z^{2}+14z+36=0 \therefore x = \frac{14\pm\sqrt{14^{2}4\cdot1\cdot36}}{2\cdot1}\end{align*}
 \begin{align*}x = 7\pm\sqrt{13}\end{align*}
Practice
 List three ways to write a quadratic function.
Solve with the quadratic formula:
 \begin{align*}m^2  5m  14 = 0\end{align*}
 \begin{align*}b^2  4b + 4 = 0\end{align*}
 \begin{align*}4b^2 + 8b + 7 = 0\end{align*}
 \begin{align*}2m^2  7m  13 = 10\end{align*}
 \begin{align*}5r^2 = 80\end{align*}
 \begin{align*}k^2  31  2k = 6  3k^2  2k\end{align*}
 If you have an equation with a power of 4, explain how you could solve it using the quadratic formula.
Solve by completing the square:
 \begin{align*}2x^2  12x + 26 = 10\end{align*}
 \begin{align*}x^2  12x + 29 = 3\end{align*}
 \begin{align*}7x^2  14x 64 = 8\end{align*}
Solve by the most efficient method:
 \begin{align*}x^4 + 13x^2 + 36 = 0\end{align*}
 \begin{align*}x^4 + 16x^2 225 = 0\end{align*}
 \begin{align*}\frac{1}{4}x^2  \frac{1}{3}x + 1 = 0\end{align*}
 \begin{align*}\frac{2}{7}c^2  \frac{1}{2}c  \frac{3}{14} = 0\end{align*}
In the quadratic formula \begin{align*} b^2  4ac\end{align*} is called the discriminant. The values of the discriminant tell us the nature of the solutions or roots of a quadratic equation, \begin{align*}ax^2 + bx + c = 0\end{align*}
 What value(s) of the discriminant result in two unique real solutions?
 What value(s) of the discriminant result in one unique real solutions?
 What value(s) of the discriminant result in two uniqueimaginary solutions?