## Intermediate Algebra: Connecting Concepts through Application

$\color{blue}{\left\{-10, 10\right\}}$
Subtract $150$ to both sides of the equation to obtain: $x^2+50-150=150-150 \\x^2-100=0$ Since $100=10^2$, the given equation can be written as: $x^2-10^2=0$ RECALL: $a^2-b^2=(a-b)(a+b)$ Factor the binomial using the formula above to obtain: $(x-10)(x+10)=0$ Use the Zero-Factor Property by equating each factor to zero to obtain: $x-10=0$ or $x+10=0$ Solve each equation to obtain: $x-10=0 \\x=0+10 \\x=10$ or $x+10=0 \\x=0-10 \\x=-10$ Therefore, the solution set is $\color{blue}{\left\{-10, 10\right\}}$.