-8y=y^2+19+8y

Solution for -8y=y^2+19+8y equation:

-8y=y^2+19+8y

We move all terms to the left:

-8y-(y^2+19+8y)=0

We get rid of parentheses

-y^2-8y-8y-19=0

We add all the numbers together, and all the variables

-1y^2-16y-19=0

a = -1; b = -16; c = -19;
Δ = b2-4ac
Δ = -162-4·(-1)·(-19)
Δ = 180
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:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-6\sqrt{5}}{2*-1}=\frac{16-6\sqrt{5}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+6\sqrt{5}}{2*-1}=\frac{16+6\sqrt{5}}{-2}
$