12x^2-20x-5=0

Solution for 12x^2-20x-5=0 equation:

12x^2-20x-5=0

a = 12; b = -20; c = -5;
Δ = b2-4ac
Δ = -202-4·12·(-5)
Δ = 640
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:

$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-8\sqrt{10}}{2*12}=\frac{20-8\sqrt{10}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+8\sqrt{10}}{2*12}=\frac{20+8\sqrt{10}}{24}
$