2y+2y^2=4y^2

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

2y+2y^2=4y^2

We move all terms to the left:

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

determiningTheFunctionDomain
2y^2-4y^2+2y=0

We add all the numbers together, and all the variables

-2y^2+2y=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:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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