-2y+4=-2y(3y+16)

Solution for -2y+4=-2y(3y+16) equation:

-2y+4=-2y(3y+16)

We move all terms to the left:

-2y+4-(-2y(3y+16))=0


We calculate terms in parentheses: -(-2y(3y+16)), so:

-2y(3y+16)

We multiply parentheses

-6y^2-32y

Back to the equation:

-(-6y^2-32y)


We get rid of parentheses

6y^2+32y-2y+4=0

We add all the numbers together, and all the variables

6y^2+30y+4=0

a = 6; b = 30; c = +4;
Δ = b2-4ac
Δ = 302-4·6·4
Δ = 804
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{804}=\sqrt{4*201}=\sqrt{4}*\sqrt{201}=2\sqrt{201}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-2\sqrt{201}}{2*6}=\frac{-30-2\sqrt{201}}{12}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+2\sqrt{201}}{2*6}=\frac{-30+2\sqrt{201}}{12}
$