0=3x^2+18x-14

Solution for 0=3x^2+18x-14 equation:

0=3x^2+18x-14

We move all terms to the left:

0-(3x^2+18x-14)=0

We add all the numbers together, and all the variables

-(3x^2+18x-14)=0

We get rid of parentheses

-3x^2-18x+14=0

a = -3; b = -18; c = +14;
Δ = b2-4ac
Δ = -182-4·(-3)·14
Δ = 492
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:

$\sqrt{\Delta}=\sqrt{492}=\sqrt{4*123}=\sqrt{4}*\sqrt{123}=2\sqrt{123}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{123}}{2*-3}=\frac{18-2\sqrt{123}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{123}}{2*-3}=\frac{18+2\sqrt{123}}{-6}
$