# How do you solve x^2-3x=10 ?

##### 1 Answer
Aug 5, 2015

You could use the quadratic formula.

#### Explanation:

For a general form quadratic equation

$\textcolor{b l u e}{a {x}^{2} + b x + c = 0}$

you can use the quadratic formula to determine its two solutions

color(blue)(x_(1,2) = (_b +- sqrt(b^2 - 4ac))/(2a)

Your quadratic looks like this

${x}^{2} - 3 x = 10$, which can be rewritten as

${x}^{2} - 3 x - 10 = 0$.

In your case, you have $a = 1$, $b = - 3$, and $c = - 10$. This means that the two solutions will be

${x}_{1 , 2} = \frac{- \left(- 3\right) \pm \sqrt{{\left(- 3\right)}^{2} - 4 \cdot 1 \cdot \left(- 10\right)}}{2 \cdot 1}$

${x}_{1 , 2} = \frac{3 \pm \sqrt{49}}{2}$

${x}_{1 , 2} = \frac{3 \pm 7}{2} = \left\{\begin{matrix}{x}_{1} = \frac{3 + 7}{2} = \textcolor{g r e e n}{5} \\ {x}_{2} = \frac{3 - 7}{2} = \textcolor{g r e e n}{- 2}\end{matrix}\right.$