РЕШИТЕ ПОЖАЛУЙСТА!!!!!!!!!!!

$\frac{4x^2+9x}{8}-\frac{3-x}{3}=\frac{2x^2-5}{5}\\\\\frac{4x^2+9x}{8}=\frac{2x^2-5}{5}+\frac{3-x}{3}=\frac{6x^2-15}{15}+\frac{15-5x}{15}=\frac{6x^2-5x}{15}\to\\\frac{4x^2+9x}{8}=\frac{6x^2-5x}{15}\to(4x^2+9x)15=8(6x^2-5x)\\\\60x^2+135x=48x^2-40x\\12x^2+175x=0\\x(12x+175)=0$
$x=0$ или $12x+175=0\to x=-\frac{175}{12}$

1. $0+(-\frac{175}{12})=-\frac{175}{12};$
2. $0*(-\frac{175}{12})=0$