An algebra problem by frame calaite

$(x+1)(x+2)=2x+4$

$x^2+2x+x+2=2x+4$

$x^2+x-2=0$

Using the quadratic formula, we have

$x= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}=\dfrac{-1 \pm \sqrt{1-4(-2)}}{-2}=- \dfrac{1}{2} \pm \dfrac{3}{2}$

Positive value: $x=- \dfrac{1}{2} + \dfrac{3}{2}=1$

Negative value: $x=- \dfrac{1}{2} - \dfrac{3}{2} = -2$

The desired answer is $\boxed{-2,1}$ .