3y2+4y=4

Solution for 3y2+4y=4 equation:

3y^2+4y=4

We move all terms to the left:

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