## College Algebra (10th Edition)

$\{-3-2\sqrt{2},-3+2\sqrt{2}\}$
STEP 1. Identify the coefficients $a=1, b=6, c=1$ STEP 2. Check the discriminant for the type and number of solutions. $b^{2}-4ac=36-4(1)(1)=32\gt 0$ so there are two unequal real solutions. STEP 3. Apply the quadratic formula $x=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ $x=\displaystyle \frac{-6\pm\sqrt{32}}{2(1)}=\frac{-6\pm\sqrt{16(2)}}{2(1)}$ $=\displaystyle \frac{-6\pm 4\sqrt{2}}{2}$ $=\displaystyle \frac{2(-3\pm 2\sqrt{2})}{2}$ $=-3\pm 2\sqrt{2}$ Solution set = $\{-3-2\sqrt{2},-3+2\sqrt{2}\}$