(x^2-5x+6)/(x^2+3x-18)=0

Solution for (x^2-5x+6)/(x^2+3x-18)=0 equation:

(x^2-5x+6)/(x^2+3x-18)=0


Domain of the equation: (x^2+3x-18)!=0

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

x^2+3x!=18

x∈R


We multiply all the terms by the denominator


(x^2-5x+6)=0

We get rid of parentheses

x^2-5x+6=0

a = 1; b = -5; c = +6;
Δ = b2-4ac
Δ = -52-4·1·6
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-1}{2*1}=\frac{4}{2}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+1}{2*1}=\frac{6}{2}
=3
$