(x^2+x-3)/(x^2*(1-x)^2)=0

Solution for (x^2+x-3)/(x^2*(1-x)^2)=0 equation:

(x^2+x-3)/(x^2(1-x)^2)=0


Domain of the equation: (x^2(1-x)^2)!=0

x∈R


We add all the numbers together, and all the variables

(x^2+x-3)/(x^2(-1x+1)^2)=0

We multiply all the terms by the denominator


(x^2+x-3)=0

We get rid of parentheses

x^2+x-3=0

a = 1; b = 1; c = -3;
Δ = b2-4ac
Δ = 12-4·1·(-3)
Δ = 13
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{13}}{2*1}=\frac{-1-\sqrt{13}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{13}}{2*1}=\frac{-1+\sqrt{13}}{2}
$