Chapter 1 - Section 1.4 - Radical Equations; Equations Quadratic in Form; Factorable Equations - 1.4 Assess Your Understanding - Page 118: 88

$x=\displaystyle \frac{5}{3}$

We solve by factoring with grouping: $3x^{3}+12x=5x^{2}+20$ $3x^{3}+12x-5x^{2}-20=0$ $3x^{3}-5x^{2}+12x-20=0$ $x^{2}(3x-5)+4(3x-5)=0$ $(3x-5)(x^{2}+4)=0$ $3x-5=0$ or $x^{2}+4=0$ $3x=5$ or $x^{2}=-4$ $x=\displaystyle \frac{5}{3}$ or $x=\pm \sqrt{-4}=$ not real $x=\displaystyle \frac{5}{3}$