The required values of x are x = {-2,  $\rm \frac{3}{2}$}.

Given

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

$\rm or, 2x^{2} + (4 - 3) x - 6 = 0$

$\rm or, 2x^{2} + 4x - 3x - 6 = 0$

$\rm or, 2x (x + 2) - 3(x + 2) = 0$

$\rm or, (2x - 3)(x + 2) = 0$

Either

$\rm (2x - 3) = 0$

$\rm or, 2x = 3$

$\rm \therefore x = \frac{3}{2}$

Or

$\rm (x + 2) = 0$

$\rm \therefore x = -2$