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The quadratic formula allows us to solve any quadratic equation that's in the form ax^2 + bx + c = 0. This article reviews how to apply the formula.

## What is the quadratic formula?

$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
$a{x}^{2}+bx+c=0$

### Example

We're given an equation and asked to solve for $q$:
$0=-7{q}^{2}+2q+9$
This equation is already in the form $a{x}^{2}+bx+c=0$, so we can apply the quadratic formula where $a=-7,b=2,c=9$:
Let's check both solutions to be sure it worked:
$q=-1$$q=\frac{9}{7}$
$\begin{array}{rl}0& =-7{q}^{2}+2q+9\\ \\ 0& =-7\left(-1{\right)}^{2}+2\left(-1\right)+9\\ \\ 0& =-7\left(1\right)-2+9\\ \\ 0& =-7-2+9\\ \\ 0& =0\end{array}$$\begin{array}{rl}0& =-7{q}^{2}+2q+9\\ \\ 0& =-7{\left(\frac{9}{7}\right)}^{2}+2\left(\frac{9}{7}\right)+9\\ \\ 0& =-7\left(\frac{81}{49}\right)+\left(\frac{18}{7}\right)+9\\ \\ 0& =-\left(\frac{81}{7}\right)+\left(\frac{18}{7}\right)+9\\ \\ 0& =-\left(\frac{63}{7}\right)+9\\ \\ 0& =-9+9\\ \\ 0& =0\end{array}$
Yep, both solutions check out.
Practice
Solve for $x$.
$-4+x+7{x}^{2}=0$

Want more practice? Check out this exercise.

## Want to join the conversation?

• Sal, How does the quadratic formula relate to business and economics?
• It helps in lots of ways. It can possibly predict the future path of certain things, especially if your graph is exponential.
• what if the equation doesn't equal zero
• then just subtract the non-zero number from the RHS to the LHS and make the RHS equal to zero.
• are there any shortcuts or patterns we can use to make calculation quicker?
• Not insofar as I know. The quadratic formula is the shortcut, unless you prefer grouping or something
• When there are two numbers in the numerator before the square root, how do I solve the problem?
• Can you show an example of what you mean?
• Can i have exra work.
• ok solve this: x^2 + 69x + 420 = 0
• I do not understand this. Can someone please explain
(1 vote)
• This is a formula, so if you can get the right numbers, you plug them into the formula and calculate the answer(s). We always have to start with a quadratic in standard form: ax^2+bx+c=0. Making one up, 3x^2+2x-5=0, we see a=3, b=2, c=-5. I teach my students to start with the discriminant, b^2-4ac. Also, especially in the beginning, put the b value in parentheses so that you square a negative number if b is negative. In our example, this gives (2)^2-4(3)(-5) = 4+60=64. If I take √64 = 8. Filling out the formula, we get x=(-2±8)/(2(3)) or breaking it into the two parts x=(-2+8)/6=1 and (-2-8)/6=-10/6=-5/2. Where is the confusion? It is always hard to answer when we cannot figure out what you do understand and where you are confused.
• What happens when the discriminant is a negative number? If it is negative would your answer be imaginary?