48=2y^2+4y

Solution for 48=2y^2+4y equation:

48=2y^2+4y

We move all terms to the left:

48-(2y^2+4y)=0

We get rid of parentheses

-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{-16}{-4}
=+4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+20}{2*-2}=\frac{24}{-4}
=-6
$