5x^*X-36x+41=0

Solution for 5x^*X-36x+41=0 equation:

5x^*x-36x+41=0

We add all the numbers together, and all the variables

-36x+5x^*x+41=0

Wy multiply elements

5x^2-36x+41=0

a = 5; b = -36; c = +41;
Δ = b2-4ac
Δ = -362-4·5·41
Δ = 476
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:

$\sqrt{\Delta}=\sqrt{476}=\sqrt{4*119}=\sqrt{4}*\sqrt{119}=2\sqrt{119}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-2\sqrt{119}}{2*5}=\frac{36-2\sqrt{119}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+2\sqrt{119}}{2*5}=\frac{36+2\sqrt{119}}{10}
$