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 = 180
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:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6\sqrt{5}}{2*12}=\frac{-18-6\sqrt{5}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6\sqrt{5}}{2*12}=\frac{-18+6\sqrt{5}}{24}
$