# How do you solve using the quadratic formula x^2-3x=-10?

Mar 9, 2018

See a solution process below:

#### Explanation:

First, add $\textcolor{red}{10}$ to each side of the equation to put the equation in standard form:

${x}^{2} - 3 x + \textcolor{red}{10} = - 10 + \textcolor{red}{10}$

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

We can use the quadratic equation to solve this problem:

For $\textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c} = 0$, the values of $x$ which are the solutions to the equation are given by:

$x = \frac{- \textcolor{b l u e}{b} \pm \sqrt{{\textcolor{b l u e}{b}}^{2} - \left(4 \textcolor{red}{a} \textcolor{g r e e n}{c}\right)}}{2 \cdot \textcolor{red}{a}}$

Substituting:

$\textcolor{red}{1}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{- 3}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{10}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{- 3} \pm \sqrt{{\textcolor{b l u e}{- 3}}^{2} - \left(4 \cdot \textcolor{red}{1} \cdot \textcolor{g r e e n}{10}\right)}}{2 \cdot \textcolor{red}{1}}$

$x = \frac{3 \pm \sqrt{9 - 40}}{2}$

$x = \frac{3 \pm \sqrt{- 31}}{2}$