2x+2x^2+4=4

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

2x+2x^2+4=4

We move all terms to the left:

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

We add all the numbers together, and all the variables

2x^2+2x=0

a = 2; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·2·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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