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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((Y+1)(Y+1)-3), so:

(Y+1)(Y+1)-3

We multiply parentheses ..

(+Y^2+Y+Y+1)-3

We get rid of parentheses

Y^2+Y+Y+1-3

We add all the numbers together, and all the variables

Y^2+2Y-2

Back to the equation:

-(Y^2+2Y-2)


We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -1; b = -4; c = +6;
Δ = b2-4ac
Δ = -42-4·(-1)·6
Δ = 40
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{10}}{2*-1}=\frac{4-2\sqrt{10}}{-2}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{10}}{2*-1}=\frac{4+2\sqrt{10}}{-2}
$