2y+8=y^2

Solution for 2y+8=y^2 equation:

2y+8=y^2

We move all terms to the left:

2y+8-(y^2)=0

determiningTheFunctionDomain
-y^2+2y+8=0

We add all the numbers together, and all the variables

-1y^2+2y+8=0

a = -1; b = 2; c = +8;
Δ = b2-4ac
Δ = 22-4·(-1)·8
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6}{2*-1}=\frac{-8}{-2}
=+4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6}{2*-1}=\frac{4}{-2}
=-2
$