solve $x^{2}-24x = - 80$ by completing the square. what is the solution set of the equation?\n{2, 40}\n{4…

Explanation:

Step1: Rewrite the equation

Given $x^{2}-24x=-80$. The coefficient of $x$ is $- 24$. Half of it is $\frac{-24}{2}=-12$, and its square is $(-12)^{2} = 144$. Add 144 to both sides of the equation: $x^{2}-24x + 144=-80 + 144$.

Step2: Factor the left - hand side

The left - hand side is a perfect square trinomial. Using the formula $(a - b)^2=a^{2}-2ab + b^{2}$, where $a = x$ and $b = 12$, we have $(x - 12)^{2}=64$.

Step3: Take the square root of both sides

$x-12=\pm\sqrt{64}=\pm8$.

Step4: Solve for $x$

When $x - 12=8$, then $x=8 + 12=20$. When $x - 12=-8$, then $x=-8 + 12 = 4$.

Answer:

B. ${4,20}$