20y^2=20y+48=(y+2)^2

Solution for 20y^2=20y+48=(y+2)^2 equation:

20y^2=20y+48=(y+2)^2

We move all terms to the left:

20y^2-(20y+48)=0

We get rid of parentheses

20y^2-20y-48=0

a = 20; b = -20; c = -48;
Δ = b2-4ac
Δ = -202-4·20·(-48)
Δ = 4240
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:

$\sqrt{\Delta}=\sqrt{4240}=\sqrt{16*265}=\sqrt{16}*\sqrt{265}=4\sqrt{265}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{265}}{2*20}=\frac{20-4\sqrt{265}}{40}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{265}}{2*20}=\frac{20+4\sqrt{265}}{40}
$