Answer
{$2,6$}
Work Step by Step
Step 1: Comparing $x^{2}-8x+12=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we obtain:
$a=1$, $b=-8$ and $c=12$
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:
$x=\frac{-(-8) \pm \sqrt {(-8)^{2}-4(1)(12)}}{2(1)}$
Step 4: $x=\frac{8 \pm \sqrt {64-48}}{2}$
Step 5: $x=\frac{8 \pm \sqrt {16}}{2}$
Step 6: $x=\frac{8 \pm 4}{2}$
Step 7: $x=\frac{8+4}{2}$ or $x=\frac{8-4}{2}$
Step 8: $x=\frac{12}{2}$ or $x=\frac{4}{2}$
Step 9: $x=6$ or $x=2$
Step 10: Therefore, the solution set is {$2,6$}.
