12y^2-48y+48+4y-16=0

Solution for 12y^2-48y+48+4y-16=0 equation:

12y^2-48y+48+4y-16=0

We add all the numbers together, and all the variables

12y^2-44y+32=0

a = 12; b = -44; c = +32;
Δ = b2-4ac
Δ = -442-4·12·32
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-44)-20}{2*12}=\frac{24}{24}
=1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-44)+20}{2*12}=\frac{64}{24}
=2+2/3
$