Question

$frac{3x^2+x}{4} - frac{2-7x}{5} = frac{3x^2+17}{10}$

Answer 1

$frac{3x^2+x}{4}-frac{2-7x}{5}=frac{3x^2+17}{10} |*20\5(3x^2+x)-4(2-7x)=2(3x^2+17)\15x^2+5x-8+28x=6x^2+34\9x^2+33x-42=0 |:3\3x^2+11x-14=0\D=121+168=289=17^2\boxed{x_1=frac{-11+17}{3*2}=-1}\boxed{x_2=frac{-11-17}{3*2}=-frac{28}{6}=-frac{14}{3}}$