Question

Solve for d: $3d^2+17d-20=0$

Answer 1

Given,

$3d^2+17d-20=0$

$\Rightarrow 3d^2+17d-20=0$

$\Rightarrow 3d^2-3d+20d-20=0$

$\Rightarrow 3d(d-1)+20(d-1)=0$

$\Rightarrow (d-1)(3d+20)=0$

$\therefore d=1,-\frac{20}{3}$