Answer
$x=1$ or $x=-3$
Work Step by Step
If $|x| = c$, then $x = c$ or $x = -c. (c>0)$
$x^2+2x=3$ (equation 1) or $x^2+2x=-3$ (equation 2)
Solve using the quadratic formula
$x=\frac{−b±\sqrt{b^2−4ac}}{2a}$
Equation 1: $x^2+2x=3$
Subtract $3$ to both sides:
$x^2+2x-3=3-3$
$x^2+2x-3=0$
$a=1$, $b=2$, $c=-3$
$x=\frac{−2±\sqrt{2^2−(4⋅1⋅-3)}}{2⋅1}$
$x=\frac{−2±\sqrt{4−(-12)}}{2}$
$x=\frac{−2±\sqrt{16}}{2}$
$x=\frac{−2±4}{2}$
$x=1$ or $x=-3$
Equation 2:
$x^2+2x=-3$
Add $3$ to both sides:
$x^2+2x+3=-3+3$
$x^2+2x+3=0$
$a=1$, $b=2$, $c=3$
$x=\frac{−2±\sqrt{2^2−(4⋅1⋅3)}}{2⋅1}$
$x=\frac{−2±\sqrt{4−(12)}}{2}$
$x=\frac{−2±\sqrt{-8}}{2}$ (no real solution since the discriminant is negative)
