(x^2-x-1)/((x^2-3x)^0.5)=0

Solution for (x^2-x-1)/((x^2-3x)^0.5)=0 equation:

(x^2-x-1)/((x^2-3x)^0.5)=0


Domain of the equation: ((x^2-3x)^0.5)!=0

x∈R


We multiply all the terms by the denominator


(x^2-x-1)=0

We get rid of parentheses

x^2-x-1=0

We add all the numbers together, and all the variables

x^2-1x-1=0

a = 1; b = -1; c = -1;
Δ = b2-4ac
Δ = -12-4·1·(-1)
Δ = 5
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{-(-1)-\sqrt{5}}{2*1}=\frac{1-\sqrt{5}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{5}}{2*1}=\frac{1+\sqrt{5}}{2}
$