3y^2+12y+4=4

Solution for 3y^2+12y+4=4 equation:

3y^2+12y+4=4

We move all terms to the left:

3y^2+12y+4-(4)=0

We add all the numbers together, and all the variables

3y^2+12y=0

a = 3; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·3·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*3}=\frac{-24}{6}
=-4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*3}=\frac{0}{6}
=0
$