Answer
$\{-3-2\sqrt{2},-3+2\sqrt{2}\}$
Work Step by Step
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 = $\{-3-2\sqrt{2},-3+2\sqrt{2}\}$
