Question:
$6 x^{2}+x-12=0$
Solution:
Given:
$6 x^{2}+x-12=0$
$\Rightarrow 6 x^{2}+9 x-8 x-12=0$
$\Rightarrow 3 x(2 x+3)-4(2 x+3)=0$
$\Rightarrow(3 x-4)(2 x+3)=0$
$\Rightarrow 3 x-4=0$ or $2 x+3=0$
$\Rightarrow x=\frac{4}{3}$ or $x=\frac{-3}{2}$
Hence, $\frac{4}{3}$ and $\frac{-3}{2}$ are the roots of the equation $6 x^{2}+x-12=0$.