4y^2-13=y^2=14

Solution for 4y^2-13=y^2=14 equation:

4y^2-13=y^2=14

We move all terms to the left:

4y^2-13-(y^2)=0

We add all the numbers together, and all the variables

3y^2-13=0

a = 3; b = 0; c = -13;
Δ = b2-4ac
Δ = 02-4·3·(-13)
Δ = 156
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:

$\sqrt{\Delta}=\sqrt{156}=\sqrt{4*39}=\sqrt{4}*\sqrt{39}=2\sqrt{39}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{39}}{2*3}=\frac{0-2\sqrt{39}}{6}
=-\frac{2\sqrt{39}}{6}
=-\frac{\sqrt{39}}{3}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{39}}{2*3}=\frac{0+2\sqrt{39}}{6}
=\frac{2\sqrt{39}}{6}
=\frac{\sqrt{39}}{3}
$