12x^2-16x-5=0

Solution for 12x^2-16x-5=0 equation:

12x^2-16x-5=0

a = 12; b = -16; c = -5;
Δ = b2-4ac
Δ = -162-4·12·(-5)
Δ = 496
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:

$\sqrt{\Delta}=\sqrt{496}=\sqrt{16*31}=\sqrt{16}*\sqrt{31}=4\sqrt{31}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{31}}{2*12}=\frac{16-4\sqrt{31}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{31}}{2*12}=\frac{16+4\sqrt{31}}{24}
$