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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We multiply parentheses ..

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-1Y^2-1Y=0

a = -1; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·(-1)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*-1}=\frac{0}{-2}
=0
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*-1}=\frac{2}{-2}
=-1
$