Chapter 5 Polynomials and Polynomial Functions - 5.4 Factor and Solve Polynomial Equations - 5.4 Exercises - Skill Practice - Page 357: 53

$x=5$

According to the exercise our equation is: $(2x-5)^2\pi(3x)\frac{1}{3}=125\pi\\(4x^2-20x+25)(x)=125\\4x^3-20x^2+25x-125=0\\(x-5)(4x^2+25)=0$ Thus $4x^2+25=0$ (but this is impossible since $x^2$ is non-negative ) or $x-5=0\\x=5$