12y=8y^2+4

Solution for 12y=8y^2+4 equation:

12y=8y^2+4

We move all terms to the left:

12y-(8y^2+4)=0

We get rid of parentheses

-8y^2+12y-4=0

a = -8; b = 12; c = -4;
Δ = b2-4ac
Δ = 122-4·(-8)·(-4)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4}{2*-8}=\frac{-16}{-16}
=1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4}{2*-8}=\frac{-8}{-16}
=1/2
$