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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

-(23x^2+183x-14)=0

We get rid of parentheses

-23x^2-183x+14=0

a = -23; b = -183; c = +14;
Δ = b2-4ac
Δ = -1832-4·(-23)·14
Δ = 34777
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-183)-\sqrt{34777}}{2*-23}=\frac{183-\sqrt{34777}}{-46}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-183)+\sqrt{34777}}{2*-23}=\frac{183+\sqrt{34777}}{-46}
$