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

Aug 29, 2016

The solution set is $\left\{1 , 0 , - 3 , - 4\right\}$

#### Explanation:

Case 1: The absolute value is positive

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

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

${x}^{2} + 3 x - 4 = 0$

$\left(x + 4\right) \left(x - 1\right) = 0$

$x = - 4 \mathmr{and} 1$

Case 2: The absolute value is negative

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

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

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

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

$- x \left(x + 3\right) = 0$

$x = 0 \mathmr{and} - 3$

Check all the solutions in the original equation. You will find that all solutions satisfy the equation.

Hopefully this helps!