41x^2+100x-3500=0

Solution for 41x^2+100x-3500=0 equation:

41x^2+100x-3500=0

a = 41; b = 100; c = -3500;
Δ = b2-4ac
Δ = 1002-4·41·(-3500)
Δ = 584000
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:

$\sqrt{\Delta}=\sqrt{584000}=\sqrt{1600*365}=\sqrt{1600}*\sqrt{365}=40\sqrt{365}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-40\sqrt{365}}{2*41}=\frac{-100-40\sqrt{365}}{82}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+40\sqrt{365}}{2*41}=\frac{-100+40\sqrt{365}}{82}
$