3x^2-26x+41=0

Solution for 3x^2-26x+41=0 equation:

3x^2-26x+41=0

a = 3; b = -26; c = +41;
Δ = b2-4ac
Δ = -262-4·3·41
Δ = 184
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:

$\sqrt{\Delta}=\sqrt{184}=\sqrt{4*46}=\sqrt{4}*\sqrt{46}=2\sqrt{46}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-2\sqrt{46}}{2*3}=\frac{26-2\sqrt{46}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+2\sqrt{46}}{2*3}=\frac{26+2\sqrt{46}}{6}
$