41x^2+8x-1024=0

Solution for 41x^2+8x-1024=0 equation:

41x^2+8x-1024=0

a = 41; b = 8; c = -1024;
Δ = b2-4ac
Δ = 82-4·41·(-1024)
Δ = 168000
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:

$\sqrt{\Delta}=\sqrt{168000}=\sqrt{1600*105}=\sqrt{1600}*\sqrt{105}=40\sqrt{105}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-40\sqrt{105}}{2*41}=\frac{-8-40\sqrt{105}}{82}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+40\sqrt{105}}{2*41}=\frac{-8+40\sqrt{105}}{82}
$