Answer
{$-4-\sqrt {13},-4+\sqrt {13}$}
Work Step by Step
Step 1: Comparing $x^{2}+8x+3=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we find:
$a=1$, $b=8$ and $c=3$
Step 2: The quadratic formula is:
$x=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$
Step 3: Substituting the values of a, b and c in the formula:
$x=\frac{-(8) \pm \sqrt {(8)^{2}-4(1)(3)}}{2(1)}$
Step 4: $x=\frac{-8 \pm \sqrt {64-12}}{2}$
Step 5: $x=\frac{-8 \pm \sqrt {52}}{2}$
Step 6: $x=\frac{-8 \pm \sqrt {4\times13}}{2}$
Step 7: $x=\frac{-8 \pm (\sqrt {4}\times\sqrt {13})}{2}$
Step 8: $x=\frac{-8 \pm (2\times \sqrt {13})}{2}$
Step 9: $x=\frac{2(-4 \pm \sqrt {13})}{2}$
Step 10: $x=-4 \pm \sqrt {13}$
Step 11: $x=-4 - \sqrt {13}$ or $x=-4 + \sqrt {13}$
Step 12: Therefore, the solution set is {$-4-\sqrt {13},-4+\sqrt {13}$}.
