Answer
{$-10.44,1.44$}
Work Step by Step
Step 1: Comparing $x^{2}+9x-15=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we obtain:
$a=1$, $b=9$, and $c=-15$
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:
$x=\frac{-(9) \pm \sqrt {(9)^{2}-4(1)(-15)}}{2(1)}$
Step 4: $x=\frac{-9 \pm \sqrt {81+60}}{2}$
Step 5: $x=\frac{-9 \pm \sqrt {141}}{2}$
Step 6: $x=\frac{-9 \pm 11.874}{2}$
Step 7: $x=\frac{-9+11.874}{2}$ or $x=\frac{-9-11.874}{2}$
Step 8: $x=\frac{2.874}{2}$ or $x=\frac{-20.874}{2}$
Step 9: $x=1.437\approx1.44$ or $x=-10.437\approx-10.44$
Step 10: Therefore, the solution set is {$-10.44,1.44$}.
