4y(y-10)+20=180

Solution for 4y(y-10)+20=180 equation:

4y(y-10)+20=180

We move all terms to the left:

4y(y-10)+20-(180)=0

We add all the numbers together, and all the variables

4y(y-10)-160=0

We multiply parentheses

4y^2-40y-160=0

a = 4; b = -40; c = -160;
Δ = b2-4ac
Δ = -402-4·4·(-160)
Δ = 4160
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:

$\sqrt{\Delta}=\sqrt{4160}=\sqrt{64*65}=\sqrt{64}*\sqrt{65}=8\sqrt{65}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-8\sqrt{65}}{2*4}=\frac{40-8\sqrt{65}}{8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+8\sqrt{65}}{2*4}=\frac{40+8\sqrt{65}}{8}
$