2y^2+16y=12y+48

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

2y^2+16y=12y+48

We move all terms to the left:

2y^2+16y-(12y+48)=0

We get rid of parentheses

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

We add all the numbers together, and all the variables

2y^2+4y-48=0

a = 2; b = 4; c = -48;
Δ = b2-4ac
Δ = 42-4·2·(-48)
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{-(4)-20}{2*2}=\frac{-24}{4}
=-6
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+20}{2*2}=\frac{16}{4}
=4
$