12x^2-18x-3=0

Solution for 12x^2-18x-3=0 equation:

12x^2-18x-3=0

a = 12; b = -18; c = -3;
Δ = b2-4ac
Δ = -182-4·12·(-3)
Δ = 468
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:

$\sqrt{\Delta}=\sqrt{468}=\sqrt{36*13}=\sqrt{36}*\sqrt{13}=6\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6\sqrt{13}}{2*12}=\frac{18-6\sqrt{13}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6\sqrt{13}}{2*12}=\frac{18+6\sqrt{13}}{24}
$