(-x^2+6x)/(x^2-x+3)^2=0

Solution for (-x^2+6x)/(x^2-x+3)^2=0 equation:

(-x^2+6x)/(x^2-x+3)^2=0


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

x∈R


We multiply all the terms by the denominator


(-x^2+6x)=0

We get rid of parentheses

-x^2+6x=0

We add all the numbers together, and all the variables

-1x^2+6x=0

a = -1; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·(-1)·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*-1}=\frac{-12}{-2}
=+6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*-1}=\frac{0}{-2}
=0
$