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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We multiply parentheses

-4y^2-3y-12y-(2(y+3))=0


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

2(y+3)

We multiply parentheses

2y+6

Back to the equation:

-(2y+6)


We add all the numbers together, and all the variables

-4y^2-15y-(2y+6)=0

We get rid of parentheses

-4y^2-15y-2y-6=0

We add all the numbers together, and all the variables

-4y^2-17y-6=0

a = -4; b = -17; c = -6;
Δ = b2-4ac
Δ = -172-4·(-4)·(-6)
Δ = 193
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-\sqrt{193}}{2*-4}=\frac{17-\sqrt{193}}{-8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+\sqrt{193}}{2*-4}=\frac{17+\sqrt{193}}{-8}
$