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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

-6y^2-8y-(-2(y+4))=0


We calculate terms in parentheses: -(-2(y+4)), so:

-2(y+4)

We multiply parentheses

-2y-8

Back to the equation:

-(-2y-8)


We get rid of parentheses

-6y^2-8y+2y+8=0

We add all the numbers together, and all the variables

-6y^2-6y+8=0

a = -6; b = -6; c = +8;
Δ = b2-4ac
Δ = -62-4·(-6)·8
Δ = 228
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:

$\sqrt{\Delta}=\sqrt{228}=\sqrt{4*57}=\sqrt{4}*\sqrt{57}=2\sqrt{57}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{57}}{2*-6}=\frac{6-2\sqrt{57}}{-12}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{57}}{2*-6}=\frac{6+2\sqrt{57}}{-12}
$