(-y^2+2y)/(y^2-3y+3)^2=0

Solution for (-y^2+2y)/(y^2-3y+3)^2=0 equation:

(-y^2+2y)/(y^2-3y+3)^2=0


Domain of the equation: (y^2-3y+3)^2!=0

y∈R


We multiply all the terms by the denominator


(-y^2+2y)=0

We get rid of parentheses

-y^2+2y=0

We add all the numbers together, and all the variables

-1y^2+2y=0

a = -1; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·(-1)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*-1}=\frac{-4}{-2}
=+2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*-1}=\frac{0}{-2}
=0
$