solve $x^{2}+10x = 24$ by completing the square. which is the solution set of the equation?\n{-5…

Explanation:

Step1: Rewrite the equation

Given $x^{2}+10x = 24$. The coefficient of $x$ is $10$. Half of it is $\frac{10}{2}=5$, and its square is $5^{2}=25$. Add 25 to both sides of the equation: $x^{2}+10x + 25=24 + 25$.

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 $b = 5$, we have $(x + 5)^{2}=49$.

Step3: Take the square root of both sides

$x + 5=\pm\sqrt{49}=\pm7$.

Step4: Solve for $x$

Case 1: When $x + 5=7$, then $x=7 - 5=2$. Case 2: When $x + 5=-7$, then $x=-7 - 5=-12$.

Answer:

{-12, 2}