A=2x^2+4x

Solution for A=2x^2+4x equation:

=2A^2+4A

We move all terms to the left:

-(2A^2+4A)=0

We get rid of parentheses

-2A^2-4A=0

a = -2; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·(-2)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*-2}=\frac{0}{-4}
=0
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*-2}=\frac{8}{-4}
=-2
$