12y^2=4=13y=0

Solution for 12y^2=4=13y=0 equation:

12y^2=4=13y=0

We move all terms to the left:

12y^2-(4)=0

a = 12; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·12·(-4)
Δ = 192
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{3}}{2*12}=\frac{0-8\sqrt{3}}{24}
=-\frac{8\sqrt{3}}{24}
=-\frac{\sqrt{3}}{3}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{3}}{2*12}=\frac{0+8\sqrt{3}}{24}
=\frac{8\sqrt{3}}{24}
=\frac{\sqrt{3}}{3}
$