Solve, using formula :

x² + x - (a + 2)(a + 1) = 0

Comparing above equation with ax² + bx + c = 0 we get,

a = 1, b = 1, c = -(a + 2)(a + 1)

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values we get,

$\Rightarrow \dfrac{-1 \pm \sqrt{(1)^2 - 4(1)(-(a + 2)(a + 1))}}{2(1)} \\[1em] = \dfrac{-1 \pm \sqrt{1 + 4(a + 2)(a + 1)}}{2} \\[1em] = \dfrac{-1 \pm \sqrt{1 + 4(a^2 + a + 2a + 2)}}{2} \\[1em] = \dfrac{-1 \pm \sqrt{1 + 4(a^2 + 3a + 2)}}{2} \\[1em] = \dfrac{-1 \pm \sqrt{4a^2 + 12a + 8 + 1}}{2} \\[1em] = \dfrac{-1 \pm \sqrt{4a^2 + 12a + 9}}{2} \\[1em] = \dfrac{-1 \pm \sqrt{(2a + 3)^2}}{2} \\[1em] = \dfrac{-1 \pm 2a + 3}{2} \\[1em] = \dfrac{-1 + (2a + 3)}{2} \text{ or } \dfrac{-1 - (2a + 3)}{2} \\[1em] = \dfrac{2a + 2}{2} \text{ or } \dfrac{-2a - 4}{2} \\[1em] = (a + 1) \text{ or } -(a + 2).$

Hence, x = (a + 1) or -(a + 2).