# How do you solve abs(x^2+6x)= 3x+18?

Jun 8, 2018

${x}_{1} = {x}_{2} = - 6$, ${x}_{3} = - 3$ and ${x}_{4} = 3$

#### Explanation:

$| {x}^{2} + 6 x | = 3 x + 18$

${\left({x}^{2} + 6 x\right)}^{2} = {\left(3 x + 18\right)}^{2}$

${\left(x \cdot \left(x + 6\right)\right)}^{2} = {\left(3 \cdot \left(x + 6\right)\right)}^{2}$

${x}^{2} \cdot {\left(x + 6\right)}^{2} = 9 \cdot {\left(x + 6\right)}^{2}$

$\left({x}^{2} - 9\right) \cdot {\left(x + 6\right)}^{2} = 0$

${\left(x + 6\right)}^{2} \left(x + 3\right) \left(x - 3\right) = 0$

So ${x}_{1} = {x}_{2} = - 6$, ${x}_{3} = - 3$ and ${x}_{4} = 3$