64x^2-40x-23=0

Solution for 64x^2-40x-23=0 equation:

64x^2-40x-23=0

a = 64; b = -40; c = -23;
Δ = b2-4ac
Δ = -402-4·64·(-23)
Δ = 7488
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:

$\sqrt{\Delta}=\sqrt{7488}=\sqrt{576*13}=\sqrt{576}*\sqrt{13}=24\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-24\sqrt{13}}{2*64}=\frac{40-24\sqrt{13}}{128}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+24\sqrt{13}}{2*64}=\frac{40+24\sqrt{13}}{128}
$