-4y=3y^2-4

Solution for -4y=3y^2-4 equation:

-4y=3y^2-4

We move all terms to the left:

-4y-(3y^2-4)=0

We get rid of parentheses

-3y^2-4y+4=0

a = -3; b = -4; c = +4;
Δ = b2-4ac
Δ = -42-4·(-3)·4
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8}{2*-3}=\frac{-4}{-6}
=2/3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*-3}=\frac{12}{-6}
=-2
$