4y^2-8y-32=28

Solution for 4y^2-8y-32=28 equation:

4y^2-8y-32=28

We move all terms to the left:

4y^2-8y-32-(28)=0

We add all the numbers together, and all the variables

4y^2-8y-60=0

a = 4; b = -8; c = -60;
Δ = b2-4ac
Δ = -82-4·4·(-60)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-32}{2*4}=\frac{-24}{8}
=-3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+32}{2*4}=\frac{40}{8}
=5
$