Answer
$x = \dfrac{1}{7}$
Work Step by Step
$x \ne {-5, -3, 2}$
$\dfrac{4}{x^2+3x-10}-\dfrac{1}{x^2+x-6} = \dfrac{3}{x^2-x-12}$
$\dfrac{4}{(x+5)(x-2)}-\dfrac{1}{(x+3)(x-2)} = \dfrac{3}{(x-4)(x+3)}$
$\dfrac{4(x+3)-(x+5)}{(x+5)(x+3)(x-2)} = \dfrac{3}{(x-4)(x+3)}$
$\dfrac{4x+12-x-5}{(x+5)(x-2)} = \dfrac{3}{(x-4)}$
$\dfrac{3x+7}{(x+5)(x-2)} = \dfrac{3}{(x-4)}$
$(3x+7)(x-4) = 3(x+5)(x-2)$
$3x^2-12x+7x-28 = 3(x^2+3x-10)$
$3x^2-5x-28 = 3x^2+9x-30$
$-5x -9x= 3x^2 - 3x^2-30+28$
$-14x = -2$
$x = \dfrac{-2}{-14}$
$x = \dfrac{1}{7}$
