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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


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

y(y+2)

We multiply parentheses

y^2+2y

Back to the equation:

-(y^2+2y)


We get rid of parentheses

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

We add all the numbers together, and all the variables

2y^2-3y-2=0

a = 2; b = -3; c = -2;
Δ = b2-4ac
Δ = -32-4·2·(-2)
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{-(-3)-5}{2*2}=\frac{-2}{4}
=-1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+5}{2*2}=\frac{8}{4}
=2
$