x^2-13x+41=0

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Solution for x^2-13x+41=0 equation: 



x^2-13x+41=0

a = 1; b = -13; c = +41;
Δ = b2-4ac
Δ = -132-4·1·41
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{-(-13)-\sqrt{5}}{2*1}=\frac{13-\sqrt{5}}{2}
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{5}}{2*1}=\frac{13+\sqrt{5}}{2}
$

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