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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-1y^2-2y+247=0

a = -1; b = -2; c = +247;
Δ = b2-4ac
Δ = -22-4·(-1)·247
Δ = 992
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:

$\sqrt{\Delta}=\sqrt{992}=\sqrt{16*62}=\sqrt{16}*\sqrt{62}=4\sqrt{62}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-4\sqrt{62}}{2*-1}=\frac{2-4\sqrt{62}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+4\sqrt{62}}{2*-1}=\frac{2+4\sqrt{62}}{-2}
$