Answer
{$-2.52,7.52$}
Work Step by Step
Step 1: Comparing $x^{2}-5x-19=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we obtain:
$a=1$, $b=-5$, and $c=-19$
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:
$x=\frac{-(-5) \pm \sqrt {(-5)^{2}-4(1)(-19)}}{2(1)}$
Step 4: $x=\frac{5 \pm \sqrt {25+76}}{2}$
Step 5: $x=\frac{5 \pm \sqrt {101}}{2}$
Step 6: $x=\frac{5 \pm 10.050}{2}$
Step 7: $x=\frac{5+10.050}{2}$ or $x=\frac{5-10.050}{2}$
Step 8: $x=\frac{15.050}{2}$ or $x=\frac{-5.050}{2}$
Step 9: $x\approx7.52$ or $x=\approx-2.52$
Step 10: Therefore, the solution set is {$-2.52,7.52$}.
