2y^2+4y=16

Solution for 2y^2+4y=16 equation:

2y^2+4y=16

We move all terms to the left:

2y^2+4y-(16)=0

a = 2; b = 4; c = -16;
Δ = b2-4ac
Δ = 42-4·2·(-16)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-12}{2*2}=\frac{-16}{4}
=-4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+12}{2*2}=\frac{8}{4}
=2
$