Answer
{$-1,6$}
Work Step by Step
Step 1: Comparing $x^{2}-5x-6=0$ to the standard form of a quadratic equation $ax^{2}+bx+c=0$, we obtain:
$a=1$, $b=-5$ and $c=-6$
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:
$x=\frac{-(-5) \pm \sqrt {(-5)^{2}-4(1)(-6)}}{2(1)}$
Step 4: $x=\frac{5 \pm \sqrt {25+24}}{2}$
Step 5: $x=\frac{5 \pm \sqrt {49}}{2}$
Step 6: $x=\frac{5 \pm 7}{2}$
Step 7: $x=\frac{5+7}{2}$ or $x=\frac{5-7}{2}$
Step 8: $x=\frac{12}{2}$ or $x=\frac{-2}{2}$
Step 9: $x=6$ or $x=-1$
Step 10: Therefore, the solution set is {$-1,6$}.
