3y(y+1)=6+7y(y+2)

Solution for 3y(y+1)=6+7y(y+2) equation:

3y(y+1)=6+7y(y+2)

We move all terms to the left:

3y(y+1)-(6+7y(y+2))=0

We multiply parentheses

3y^2+3y-(6+7y(y+2))=0


We calculate terms in parentheses: -(6+7y(y+2)), so:

6+7y(y+2)

determiningTheFunctionDomain
7y(y+2)+6

We multiply parentheses

7y^2+14y+6

Back to the equation:

-(7y^2+14y+6)


We get rid of parentheses

3y^2-7y^2+3y-14y-6=0

We add all the numbers together, and all the variables

-4y^2-11y-6=0

a = -4; b = -11; c = -6;
Δ = b2-4ac
Δ = -112-4·(-4)·(-6)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-5}{2*-4}=\frac{6}{-8}
=-3/4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+5}{2*-4}=\frac{16}{-8}
=-2
$