M(x)=-2x^2

Solution for M(x)=-2x^2 equation:

(M)=-2M^2

We move all terms to the left:

(M)-(-2M^2)=0

We get rid of parentheses

2M^2+M=0

a = 2; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·2·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*2}=\frac{-2}{4}
=-1/2
$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*2}=\frac{0}{4}
=0
$