Answer
{$-2 - 3\sqrt {2},-2 + 3\sqrt {2}$}
Work Step by Step
Step 1: Comparing $x^{2}+4x-14=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we find:
$a=1$, $b=4$ and $c=-14$
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:
$x=\frac{-(4) \pm \sqrt {(4)^{2}-4(1)(-14)}}{2(1)}$
Step 4: $x=\frac{-4 \pm \sqrt {16+56}}{2}$
Step 5: $x=\frac{-4 \pm \sqrt {72}}{2}$
Step 6: $x=\frac{-4 \pm \sqrt {36\times2}}{2}$
Step 7: $x=\frac{-4 \pm (\sqrt {36}\times\sqrt {2})}{2}$
Step 8: $x=\frac{-4 \pm (6\times \sqrt {2})}{2}$
Step 9: $x=\frac{2(-2 \pm 3\sqrt {2}}{2}$
Step 10: $x=-2 \pm 3\sqrt {2}$
Step 11: $x=-2 - 3\sqrt {2}$ or $x=-2 + 3\sqrt {2}$
Step 12: Therefore, the solution set is {$-2 - 3\sqrt {2},-2 + 3\sqrt {2}$}.
