Y2-2y=(y-2)(y+3)

Solution for Y2-2y=(y-2)(y+3) equation:

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

We move all terms to the left:

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

We multiply parentheses ..

-((+Y^2+3Y-2Y-6))-2Y+2=0


We calculate terms in parentheses: -((+Y^2+3Y-2Y-6)), so:

(+Y^2+3Y-2Y-6)

We get rid of parentheses

Y^2+3Y-2Y-6

We add all the numbers together, and all the variables

Y^2+Y-6

Back to the equation:

-(Y^2+Y-6)


We add all the numbers together, and all the variables

-2Y-(Y^2+Y-6)+2=0

We get rid of parentheses

-Y^2-2Y-Y+6+2=0

We add all the numbers together, and all the variables

-1Y^2-3Y+8=0

a = -1; b = -3; c = +8;
Δ = b2-4ac
Δ = -32-4·(-1)·8
Δ = 41
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{-(-3)-\sqrt{41}}{2*-1}=\frac{3-\sqrt{41}}{-2}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{41}}{2*-1}=\frac{3+\sqrt{41}}{-2}
$