(x^2-x-2)/(x^2+4x+3)=0

Solution for (x^2-x-2)/(x^2+4x+3)=0 equation:

(x^2-x-2)/(x^2+4x+3)=0


Domain of the equation: (x^2+4x+3)!=0

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

x^2+4x!=-3

x∈R


We multiply all the terms by the denominator


(x^2-x-2)=0

We get rid of parentheses

x^2-x-2=0

We add all the numbers together, and all the variables

x^2-1x-2=0

a = 1; b = -1; c = -2;
Δ = b2-4ac
Δ = -12-4·1·(-2)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-3}{2*1}=\frac{-2}{2}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+3}{2*1}=\frac{4}{2}
=2
$