20y^2-8+y^2=15

Solution for 20y^2-8+y^2=15 equation:

20y^2-8+y^2=15

We move all terms to the left:

20y^2-8+y^2-(15)=0

We add all the numbers together, and all the variables

21y^2-23=0

a = 21; b = 0; c = -23;
Δ = b2-4ac
Δ = 02-4·21·(-23)
Δ = 1932
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:

$\sqrt{\Delta}=\sqrt{1932}=\sqrt{4*483}=\sqrt{4}*\sqrt{483}=2\sqrt{483}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{483}}{2*21}=\frac{0-2\sqrt{483}}{42}
=-\frac{2\sqrt{483}}{42}
=-\frac{\sqrt{483}}{21}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{483}}{2*21}=\frac{0+2\sqrt{483}}{42}
=\frac{2\sqrt{483}}{42}
=\frac{\sqrt{483}}{21}
$