2y+57=y^2

Solution for 2y+57=y^2 equation:

2y+57=y^2

We move all terms to the left:

2y+57-(y^2)=0

determiningTheFunctionDomain
-y^2+2y+57=0

We add all the numbers together, and all the variables

-1y^2+2y+57=0

a = -1; b = 2; c = +57;
Δ = b2-4ac
Δ = 22-4·(-1)·57
Δ = 232
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:

$\sqrt{\Delta}=\sqrt{232}=\sqrt{4*58}=\sqrt{4}*\sqrt{58}=2\sqrt{58}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{58}}{2*-1}=\frac{-2-2\sqrt{58}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{58}}{2*-1}=\frac{-2+2\sqrt{58}}{-2}
$