4y+2y2=16+8y

Solution for 4y+2y2=16+8y equation:

4y+2y^2=16+8y

We move all terms to the left:

4y+2y^2-(16+8y)=0

We add all the numbers together, and all the variables

2y^2+4y-(8y+16)=0

We get rid of parentheses

2y^2+4y-8y-16=0

We add all the numbers together, and all the variables

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