Answer
{$-1.52,8.52$}
Work Step by Step
Step 1: Comparing $x^{2}-7x-13=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we obtain:
$a=1$, $b=-7$, and $c=-13$
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:
$x=\frac{-(-7) \pm \sqrt {(-7)^{2}-4(1)(-13)}}{2(1)}$
Step 4: $x=\frac{7 \pm \sqrt {49+52}}{2}$
Step 5: $x=\frac{7 \pm \sqrt {101}}{2}$
Step 6: $x=\frac{7 \pm 10.050}{2}$
Step 7: $x=\frac{7+10.050}{2}$ or $x=\frac{7-10.050}{2}$
Step 8: $x=\frac{17.050}{2}$ or $x=\frac{-3.050}{2}$
Step 9: $x\approx8.52$ or $x=\approx-1.52$
Step 10: Therefore, the solution set is {$-1.52,8.52$}.
