41x^2+50x-540=0

Solution for 41x^2+50x-540=0 equation:

41x^2+50x-540=0

a = 41; b = 50; c = -540;
Δ = b2-4ac
Δ = 502-4·41·(-540)
Δ = 91060
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:

$\sqrt{\Delta}=\sqrt{91060}=\sqrt{4*22765}=\sqrt{4}*\sqrt{22765}=2\sqrt{22765}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-2\sqrt{22765}}{2*41}=\frac{-50-2\sqrt{22765}}{82}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+2\sqrt{22765}}{2*41}=\frac{-50+2\sqrt{22765}}{82}
$