(2x^2-9x+3)/(x-3)=x

Solution for (2x^2-9x+3)/(x-3)=x equation:

(2x^2-9x+3)/(x-3)=x

We move all terms to the left:

(2x^2-9x+3)/(x-3)-(x)=0


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

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

x!=3

x∈R


We add all the numbers together, and all the variables

-1x+(2x^2-9x+3)/(x-3)=0

We multiply all the terms by the denominator


-1x*(x-3)+(2x^2-9x+3)=0

We multiply parentheses

-x^2+3x+(2x^2-9x+3)=0

We get rid of parentheses

-x^2+2x^2+3x-9x+3=0

We add all the numbers together, and all the variables

x^2-6x+3=0

a = 1; b = -6; c = +3;
Δ = b2-4ac
Δ = -62-4·1·3
Δ = 24
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{6}}{2*1}=\frac{6-2\sqrt{6}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{6}}{2*1}=\frac{6+2\sqrt{6}}{2}
$