18x^2+20x-12=0

Solution for 18x^2+20x-12=0 equation:

18x^2+20x-12=0

a = 18; b = 20; c = -12;
Δ = b2-4ac
Δ = 202-4·18·(-12)
Δ = 1264
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:

$\sqrt{\Delta}=\sqrt{1264}=\sqrt{16*79}=\sqrt{16}*\sqrt{79}=4\sqrt{79}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{79}}{2*18}=\frac{-20-4\sqrt{79}}{36}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{79}}{2*18}=\frac{-20+4\sqrt{79}}{36}
$