12x2+18x-4=0

Solution for 12x2+18x-4=0 equation:

12x^2+18x-4=0

a = 12; b = 18; c = -4;
Δ = b2-4ac
Δ = 182-4·12·(-4)
Δ = 516
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:

$\sqrt{\Delta}=\sqrt{516}=\sqrt{4*129}=\sqrt{4}*\sqrt{129}=2\sqrt{129}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{129}}{2*12}=\frac{-18-2\sqrt{129}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{129}}{2*12}=\frac{-18+2\sqrt{129}}{24}
$