(x^2+3x-4)/(x^2+2x-8)=0

Solution for (x^2+3x-4)/(x^2+2x-8)=0 equation:

(x^2+3x-4)/(x^2+2x-8)=0


Domain of the equation: (x^2+2x-8)!=0

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

x^2+2x!=8

x∈R


We multiply all the terms by the denominator


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

We get rid of parentheses

x^2+3x-4=0

a = 1; b = 3; c = -4;
Δ = b2-4ac
Δ = 32-4·1·(-4)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-5}{2*1}=\frac{-8}{2}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+5}{2*1}=\frac{2}{2}
=1
$