(Y+3)(y-2)=y(y-4)-1

Solution for (Y+3)(y-2)=y(y-4)-1 equation:

(+3)(Y-2)=Y(Y-4)-1

We move all terms to the left:

(+3)(Y-2)-(Y(Y-4)-1)=0

We add all the numbers together, and all the variables

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

We multiply parentheses

3Y-(Y(Y-4)-1)-6=0


We calculate terms in parentheses: -(Y(Y-4)-1), so:

Y(Y-4)-1

We multiply parentheses

Y^2-4Y-1

Back to the equation:

-(Y^2-4Y-1)


We get rid of parentheses

-Y^2+3Y+4Y+1-6=0

We add all the numbers together, and all the variables

-1Y^2+7Y-5=0

a = -1; b = 7; c = -5;
Δ = b2-4ac
Δ = 72-4·(-1)·(-5)
Δ = 29
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{-(7)-\sqrt{29}}{2*-1}=\frac{-7-\sqrt{29}}{-2}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{29}}{2*-1}=\frac{-7+\sqrt{29}}{-2}
$