5(y+1)-7=3y(y-1)+2y

Solution for 5(y+1)-7=3y(y-1)+2y equation:

5(y+1)-7=3y(y-1)+2y

We move all terms to the left:

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

We multiply parentheses

5y-(3y(y-1)+2y)+5-7=0


We calculate terms in parentheses: -(3y(y-1)+2y), so:

3y(y-1)+2y

We add all the numbers together, and all the variables

2y+3y(y-1)

We multiply parentheses

3y^2+2y-3y

We add all the numbers together, and all the variables

3y^2-1y

Back to the equation:

-(3y^2-1y)


We add all the numbers together, and all the variables

5y-(3y^2-1y)-2=0

We get rid of parentheses

-3y^2+5y+1y-2=0

We add all the numbers together, and all the variables

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

a = -3; b = 6; c = -2;
Δ = b2-4ac
Δ = 62-4·(-3)·(-2)
Δ = 12
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:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{3}}{2*-3}=\frac{-6-2\sqrt{3}}{-6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{3}}{2*-3}=\frac{-6+2\sqrt{3}}{-6}
$