40x^2+2x-129=0

Solution for 40x^2+2x-129=0 equation:

40x^2+2x-129=0

a = 40; b = 2; c = -129;
Δ = b2-4ac
Δ = 22-4·40·(-129)
Δ = 20644
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:

$\sqrt{\Delta}=\sqrt{20644}=\sqrt{4*5161}=\sqrt{4}*\sqrt{5161}=2\sqrt{5161}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{5161}}{2*40}=\frac{-2-2\sqrt{5161}}{80}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{5161}}{2*40}=\frac{-2+2\sqrt{5161}}{80}
$