# How do you solve x^2 + 18 = –9x?

Aug 21, 2015

${x}_{1 , 2} = \frac{- 9 \pm 3}{2}$

#### Explanation:

Start by getting all the terms on one side of the equation to get it classic quadratic form

${x}^{2} + 9 x + 18 = 0$

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

$\textcolor{b l u e}{{x}_{1 , 2} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}}$

In your case, you have $a = 1$, $b = 9$, and $c = 18$. This means that you have

${x}_{1 , 2} = \frac{- 9 \pm \sqrt{{9}^{2} - 4 \cdot 1 \cdot 18}}{2 \cdot 1}$

${x}_{1 , 2} = \frac{- 9 \pm \sqrt{9}}{2}$

${x}_{1 , 2} = \frac{- 9 \pm 3}{2} = \left\{\begin{matrix}{x}_{1} = \frac{- 9 - 3}{2} = \textcolor{g r e e n}{- 6} \\ {x}_{2} = \frac{- 9 + 3}{2} = \textcolor{g r e e n}{- 3}\end{matrix}\right.$

Aug 22, 2015

Solve $y = {x}^{2} + 9 x + 18 = 0$

Ans: -3 and -6

#### Explanation:

I use the new Transforming Method (Google, Yahoo Search)
Find the 2 real roots, knowing sum (-b = -9) and product (c = 18).
Both roots are negative (Rule of Signs)
Factor pairs of (18) --> (-2, -9)(-3, -6). This sum is -9 = -b. Then, the 2 real roots are -3 and -6.