Answer
{$6,9$}
Work Step by Step
Step 1: Comparing $x^{2}-15x+54=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we obtain:
$a=1$, $b=-15$, and $c=54$
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:
$x=\frac{-(-15) \pm \sqrt {(-15)^{2}-4(1)(54)}}{2(1)}$
Step 4: $x=\frac{15 \pm \sqrt {225-216}}{2}$
Step 5: $x=\frac{15 \pm \sqrt {9}}{2}$
Step 6: $x=\frac{15 \pm 3}{2}$
Step 7: $x=\frac{15+3}{2}$ or $x=\frac{15-3}{2}$
Step 8: $x=\frac{18}{2}$ or $x=\frac{12}{2}$
Step 9: $x=9$ or $x=6$
Step 10: Therefore, the solution set is {$6,9$}.
