x^2-13x+41=4

Solution for x^2-13x+41=4 equation:

x^2-13x+41=4

We move all terms to the left:

x^2-13x+41-(4)=0

We add all the numbers together, and all the variables

x^2-13x+37=0

a = 1; b = -13; c = +37;
Δ = b2-4ac
Δ = -132-4·1·37
Δ = 21
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{-(-13)-\sqrt{21}}{2*1}=\frac{13-\sqrt{21}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{21}}{2*1}=\frac{13+\sqrt{21}}{2}
$