41x^2-540x+360=0

Solution for 41x^2-540x+360=0 equation:

41x^2-540x+360=0

a = 41; b = -540; c = +360;
Δ = b2-4ac
Δ = -5402-4·41·360
Δ = 232560
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:

$\sqrt{\Delta}=\sqrt{232560}=\sqrt{144*1615}=\sqrt{144}*\sqrt{1615}=12\sqrt{1615}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-540)-12\sqrt{1615}}{2*41}=\frac{540-12\sqrt{1615}}{82}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-540)+12\sqrt{1615}}{2*41}=\frac{540+12\sqrt{1615}}{82}
$