x^2/(1-x)(2-x)=5

Solution for x^2/(1-x)(2-x)=5 equation:

x^2/(1-x)(2-x)=5

We move all terms to the left:

x^2/(1-x)(2-x)-(5)=0


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

We move all terms containing x to the left, all other terms to the right

-x)(2-x!=-1

x∈R


We add all the numbers together, and all the variables

x^2/(-1x+1)(-1x+2)-5=0

We multiply parentheses ..

x^2/(+x^2-2x-1x+2)-5=0

We multiply all the terms by the denominator


x^2-5*(+x^2-2x-1x+2)=0

We multiply parentheses

x^2-5x^2+10x+5x-10=0

We add all the numbers together, and all the variables

-4x^2+15x-10=0

a = -4; b = 15; c = -10;
Δ = b2-4ac
Δ = 152-4·(-4)·(-10)
Δ = 65
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{-(15)-\sqrt{65}}{2*-4}=\frac{-15-\sqrt{65}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{65}}{2*-4}=\frac{-15+\sqrt{65}}{-8}
$