2(2y-5)+(y^2-62)=180

Solution for 2(2y-5)+(y^2-62)=180 equation:

2(2y-5)+(y^2-62)=180

We move all terms to the left:

2(2y-5)+(y^2-62)-(180)=0

We multiply parentheses

4y+(y^2-62)-10-180=0

We get rid of parentheses

y^2+4y-62-10-180=0

We add all the numbers together, and all the variables

y^2+4y-252=0

a = 1; b = 4; c = -252;
Δ = b2-4ac
Δ = 42-4·1·(-252)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-32}{2*1}=\frac{-36}{2}
=-18
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+32}{2*1}=\frac{28}{2}
=14
$