Quadratic Equation

Let $x^2 + x = a$

So , the given equation becomes -

$(a+1)(a+2) = 12$

$a^2 + 3a + 2 - 12 = 0$

$a^2 + 3a - 10 = 0$

$(a+5)(a-2) = 0$

This gives us that $\displaystyle\ a=(-5,2)$

$When, a = 2 \Rightarrow\ x^2 + x = 2\Rightarrow\ x^2 + x -2 = 0\Rightarrow\ (x+2)(x-1)=0$

$\Rightarrow\boxed{\displaystyle\ x=(1,-2)}$

$When, a = -5 \Rightarrow\ x^2 + x = -5\Rightarrow\ x^2 + x +5 = 0$

$\Rightarrow\boxed{\displaystyle\ x=\bigg(\frac{-1+\sqrt{-19}}{2} , \frac{-1-\sqrt{-19}}{2}\bigg )}$