Y(2y-4)=(2y+1)(y-2)

Solution for Y(2y-4)=(2y+1)(y-2) equation:

(2Y-4)=(2Y+1)(Y-2)

We move all terms to the left:

(2Y-4)-((2Y+1)(Y-2))=0

We get rid of parentheses

2Y-((2Y+1)(Y-2))-4=0

We multiply parentheses ..

-((+2Y^2-4Y+Y-2))+2Y-4=0


We calculate terms in parentheses: -((+2Y^2-4Y+Y-2)), so:

(+2Y^2-4Y+Y-2)

We get rid of parentheses

2Y^2-4Y+Y-2

We add all the numbers together, and all the variables

2Y^2-3Y-2

Back to the equation:

-(2Y^2-3Y-2)


We add all the numbers together, and all the variables

2Y-(2Y^2-3Y-2)-4=0

We get rid of parentheses

-2Y^2+2Y+3Y+2-4=0

We add all the numbers together, and all the variables

-2Y^2+5Y-2=0

a = -2; b = 5; c = -2;
Δ = b2-4ac
Δ = 52-4·(-2)·(-2)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-3}{2*-2}=\frac{-8}{-4}
=+2
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+3}{2*-2}=\frac{-2}{-4}
=1/2
$