y^2-4y-8=5y-28

Solution for y^2-4y-8=5y-28 equation:

y^2-4y-8=5y-28

We move all terms to the left:

y^2-4y-8-(5y-28)=0

We get rid of parentheses

y^2-4y-5y+28-8=0

We add all the numbers together, and all the variables

y^2-9y+20=0

a = 1; b = -9; c = +20;
Δ = b2-4ac
Δ = -92-4·1·20
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-1}{2*1}=\frac{8}{2}
=4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+1}{2*1}=\frac{10}{2}
=5
$