2x^2+9x-18=x^2+3x-2

Solution for 2x^2+9x-18=x^2+3x-2 equation:

2x^2+9x-18=x^2+3x-2

We move all terms to the left:

2x^2+9x-18-(x^2+3x-2)=0

We get rid of parentheses

2x^2-x^2+9x-3x+2-18=0

We add all the numbers together, and all the variables

x^2+6x-16=0

a = 1; b = 6; c = -16;
Δ = b2-4ac
Δ = 62-4·1·(-16)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-10}{2*1}=\frac{-16}{2}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+10}{2*1}=\frac{4}{2}
=2
$