Intermediate Algebra for College Students (7th Edition)

$x=\frac{−3±\sqrt{17}}{2}$ $x=-1$ $x=-2$
If $|x| = c$, then $x = c$ or $x = -c. (c>0)$ $x^2+3x=2$ (equation 1) or $x^2+3x=-2$ (equation 2) Solve using the quadratic formula $x=\frac{−b±\sqrt{b^2−4ac}}{2a}$ Equation 1:$x^2+3x=2$ Subtract $2$ to both sides: $x^2+3x-2=2-2$ $x^2+3x-2=0$ $a=1$, $b=3$, $c=-2$ $x=\frac{−3±\sqrt{3^2−(4⋅1⋅-2)}}{2⋅1}$ $x=\frac{−3±\sqrt{9−(-8)}}{2}$ $x=\frac{−3±\sqrt{17}}{2}$ Equation 2: $x^2+3x=-2$ Add $2$ to both sides: $x^2+3x+2=-2+2$ $x^2+3x+2=0$ $a=1$, $b=3$, $c=2$ $x=\frac{−3±\sqrt{3^2−(4⋅1⋅2)}}{2⋅1}$ $x=\frac{−3±\sqrt{9−(8)}}{2}$ $x=\frac{−3±\sqrt{1}}{2}$ $x=\frac{−3+1}{2}$ or $x=\frac{−3-1}{2}$ $x=-1$ or $x=-2$