Question

Solve each of the following quadratic equations:
$6x^2+x-12=0$

Answer 1

$6x^2+x-12=06x^2+(9-8)x-12=06x^2+9x-8x-12=03x(2x+3)-4(2x+3)=0(2x+3)(3x-4)=0$

When,

$2x+3=0x=\frac{-3}{2}$

And

$3x-4=0x=\frac{4}{3}\therefore x=\frac{-3}{2}or\frac{4}{3}$