Answer
{$-9,4$}
Work Step by Step
Step 1: Comparing $x^{2}+5x-36=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we obtain:
$a=1$, $b=5$ and $c=-36$
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, we obtain:
$x=\frac{-(5) \pm \sqrt {(5)^{2}-4(1)(-36)}}{2(1)}$
Step 4: $x=\frac{-5 \pm \sqrt {25+144}}{2}$
Step 5: $x=\frac{-5 \pm \sqrt {169}}{2}$
Step 6: $x=\frac{-5 \pm 13}{2}$
Step 7: $x=\frac{-5+13}{2}$ or $x=\frac{-5-13}{2}$
Step 8: $x=\frac{8}{2}$ or $x=\frac{-18}{2}$
Step 9: $x=4$ or $x=-9$
Step 10: Therefore, the solution set is {$-9,4$}.
