Y+(3+2y)=y(2y+3)

Solution for Y+(3+2y)=y(2y+3) equation:

+(3+2Y)=Y(2Y+3)

We move all terms to the left:

+(3+2Y)-(Y(2Y+3))=0

We add all the numbers together, and all the variables

(2Y+3)-(Y(2Y+3))=0

We get rid of parentheses

2Y-(Y(2Y+3))+3=0


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

Y(2Y+3)

We multiply parentheses

2Y^2+3Y

Back to the equation:

-(2Y^2+3Y)


We get rid of parentheses

-2Y^2+2Y-3Y+3=0

We add all the numbers together, and all the variables

-2Y^2-1Y+3=0

a = -2; b = -1; c = +3;
Δ = b2-4ac
Δ = -12-4·(-2)·3
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{-(-1)-5}{2*-2}=\frac{-4}{-4}
=1
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+5}{2*-2}=\frac{6}{-4}
=-1+1/2
$