CCSS Math: HSA.REI.B.4b
Gain more insight into the quadratic formula and how it is used in quadratic equations.
The quadratic formula helps you solve quadratic equations, and is probably one of the top five formulas in math.  We’re not big fans of you memorizing formulas, but this one is useful (and we think you should learn how to derive it as well as use it, but that’s for the second video!).
If you have a general quadratic equation like this:
$ax^2+bx+c=0$
Then the formula will help you find the roots of a quadratic equation, i.e. the values of $x$ where this equation is solved.

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$
It may look a little scary, but you’ll get used to it quickly!
Practice using the formula now.

### Worked example

First we need to identify the values for a, b, and c (the coefficients). First step, make sure the equation is in the format from above, $ax^2 + bx + c = 0$:
$x^2+4x-21=0$
• $a$ is the coefficient in front of $x^2$, so here $a = 1$ (note that $a$ can’t equal $0$ -- the $x^2$ is what makes it a quadratic).
• $b$ is the coefficient in front of the $x$, so here $b = 4$.
• $c$ is the constant, or the term without any $x$ next to it, so here $c = -21$.
Then we plug $a$, $b$, and $c$ into the formula:
$x=\dfrac{-4\pm\sqrt{16-4\cdot 1\cdot (-21)}}{2}$
solving this looks like:
\begin{aligned} x&=\dfrac{-4\pm\sqrt{100}}{2} \\\\ &=\dfrac{-4\pm 10}{2} \\\\ &=-2\pm 5 \end{aligned}
Therefore $x = 3$ or $x = -7$.

## What does the solution tell us?

The two solutions are the x-intercepts of the equation, i.e. where the curve crosses the x-axis. The equation $x^2 + 3x - 4 = 0$ looks like:
where the solutions to the quadratic formula, and the intercepts are $x = -4$ and $x = 1$.
Now you can also solve a quadratic equation through factoring, completing the square, or graphing, so why do we need the formula?
Because sometimes quadratic equations are a lot harder to solve than that first example.

## Second worked example

Let’s try this for an equation that is hard to factor:
$3x^2+6x=-10$
Let’s first get it into the form where all terms are on the left-hand side:
$\underbrace{(3)}_{a}x^2+\underbrace{(6)}_{b}x+\underbrace{(10)}_{c}=0$
The formula gives us:
\begin{aligned} x&=\dfrac{-6\pm\sqrt{6^2-4\cdot 3\cdot 10}}{2\cdot 3} \\\\ &=\dfrac{-6\pm\sqrt{36-120}}{6} \\\\ &=\dfrac{-6\pm\sqrt{-84}}{6} \end{aligned}
We know you can’t take the square root of a negative number without using imaginary numbers, so that tells us there’s no real solutions to this equation.  This means that at no point will $y = 0$, the function won’t intercept the x-axis.  We can also see this when graphed on a calculator:
Quadratic equation shown on a calculator
Now you’ve got the basics of the quadratic formula!
There are many more worked examples in the videos to follow.

## Tips when using the quadratic formula

• Be careful that the equation is arranged in the right form: $ax^2 + bx + c = 0$ or it won’t work!
• Make sure you take the square root of the whole $(b^2 - 4ac)$, and that $2a$ is the denominator of everything above it
• Watch your negatives: $b^2$ can’t be negative, so if $b$ starts as negative, make sure it changes to a positive since the square of a negative or a positive is a positive
• Keep the $+/-$ and always be on the look out for TWO solutions
• If you use a calculator, the answer might be rounded to a certain number of decimal places. If asked for the exact answer (as usually happens) and the square roots can’t be easily simplified, keep the square roots in the answer, e.g. $\frac{2 - \sqrt{10}}{2}$ and $\frac{2 + \sqrt{10}}{2}$

## Next step:

• Watch Sal do an example: