(-2y/3)^2+2y+2=0

Solution for (-2y/3)^2+2y+2=0 equation:

(-2y/3)^2+2y+2=0

We add all the numbers together, and all the variables

2y+(-2y/3)^2+2=0

We multiply all the terms by the denominator


2y*3)^2+(-2y+2*3)^2=0

We add all the numbers together, and all the variables

2y*3)^2+(-2y+6)^2=0

We add all the numbers together, and all the variables

-2y+2y*3)^2+(=0

Wy multiply elements

6y^2-2y=0

a = 6; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·6·0
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*6}=\frac{0}{12}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*6}=\frac{4}{12}
=1/3
$