Solve each of the following quadratic equations:

We write, $x=4 x-3 x$ as $2 x^{2} \times(-6)=-12 x^{2}=4 x \times(-3 x)$

$\therefore 2 x^{2}+x-6=0$

$\Rightarrow 2 x^{2}+4 x-3 x-6=0$

$\Rightarrow 2 x(x+2)-3(x+2)=0$

$\Rightarrow(x+2)(2 x-3)=0$

$\Rightarrow x+2=0$ or $2 x-3=0$

$\Rightarrow x=-2$ or $x=\frac{3}{2}$

Hence, the roots of the given equation are $-2$ and $\frac{3}{2}$.