# How do you solve n^2=3n-18?

Mar 27, 2018

See a solution process below:

#### Explanation:

First, transform the equation into standard quadratic form:

${n}^{2} - \textcolor{red}{3 n} + \textcolor{b l u e}{18} = 3 n - \textcolor{red}{3 n} - 18 + \textcolor{b l u e}{18}$

${n}^{2} - 3 n + 18 = 0 - 0$

${n}^{2} - 3 n + 18 = 0$

We can now use the quadratic equation to solve this problem:

The quadratic formula states:

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 $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{18}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{- 3} \pm \sqrt{{\textcolor{b l u e}{\left(- 3\right)}}^{2} - \left(4 \cdot \textcolor{red}{1} \cdot \textcolor{g r e e n}{18}\right)}}{2 \cdot \textcolor{red}{1}}$

$x = \frac{3 \pm \sqrt{9 - 72}}{2}$

$x = \frac{3 \pm \sqrt{- 63}}{2}$