-3y(y+4)-2y=2(y+1)

Solution for -3y(y+4)-2y=2(y+1) equation:

-3y(y+4)-2y=2(y+1)

We move all terms to the left:

-3y(y+4)-2y-(2(y+1))=0

We add all the numbers together, and all the variables

-2y-3y(y+4)-(2(y+1))=0

We multiply parentheses

-3y^2-2y-12y-(2(y+1))=0


We calculate terms in parentheses: -(2(y+1)), so:

2(y+1)

We multiply parentheses

2y+2

Back to the equation:

-(2y+2)


We add all the numbers together, and all the variables

-3y^2-14y-(2y+2)=0

We get rid of parentheses

-3y^2-14y-2y-2=0

We add all the numbers together, and all the variables

-3y^2-16y-2=0

a = -3; b = -16; c = -2;
Δ = b2-4ac
Δ = -162-4·(-3)·(-2)
Δ = 232
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:

$\sqrt{\Delta}=\sqrt{232}=\sqrt{4*58}=\sqrt{4}*\sqrt{58}=2\sqrt{58}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{58}}{2*-3}=\frac{16-2\sqrt{58}}{-6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{58}}{2*-3}=\frac{16+2\sqrt{58}}{-6}
$