x^2+3x-18=-x^2+2x-6

Solution for x^2+3x-18=-x^2+2x-6 equation:

x^2+3x-18=-x^2+2x-6

We move all terms to the left:

x^2+3x-18-(-x^2+2x-6)=0

We get rid of parentheses

x^2+x^2-2x+3x+6-18=0

We add all the numbers together, and all the variables

2x^2+x-12=0

a = 2; b = 1; c = -12;
Δ = b2-4ac
Δ = 12-4·2·(-12)
Δ = 97
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{-(1)-\sqrt{97}}{2*2}=\frac{-1-\sqrt{97}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{97}}{2*2}=\frac{-1+\sqrt{97}}{4}
$