## Algebra 2 (1st Edition)

$x=5$
According to the exercise our equation is: $2x(x-1)(x-4)=40\\x(x-1)(x-4)=20\\x(x^2-5x+4)=20\\x^3-5x^2+4x-20=0\\(x-5)(x^2+4)$ Thus $x^2+4=0$ (but this is impossible since $x^2$ is non-negative) or $x-5=0\\x=5$