x^2-4x^-1=5

Solution for x^2-4x^-1=5 equation:

x^2-4x^-1=5

We move all terms to the left:

x^2-4x^-1-(5)=0

We add all the numbers together, and all the variables

x^2-4x-6=0

a = 1; b = -4; c = -6;
Δ = b2-4ac
Δ = -42-4·1·(-6)
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{10}}{2*1}=\frac{4-2\sqrt{10}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{10}}{2*1}=\frac{4+2\sqrt{10}}{2}
$