3x^2+5x-12=3x^-x+18

Solution for 3x^2+5x-12=3x^-x+18 equation:

3x^2+5x-12=3x^-x+18

We move all terms to the left:

3x^2+5x-12-(3x^-x+18)=0

We get rid of parentheses

3x^2+5x-3x^+x-18-12=0

We add all the numbers together, and all the variables

3x^2+3x-30=0

a = 3; b = 3; c = -30;
Δ = b2-4ac
Δ = 32-4·3·(-30)
Δ = 369
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:

$\sqrt{\Delta}=\sqrt{369}=\sqrt{9*41}=\sqrt{9}*\sqrt{41}=3\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{41}}{2*3}=\frac{-3-3\sqrt{41}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{41}}{2*3}=\frac{-3+3\sqrt{41}}{6}
$