3^2+2x=2x^2+3x+6

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

3^2+2x=2x^2+3x+6

We move all terms to the left:

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

We add all the numbers together, and all the variables

2x-(2x^2+3x+6)+9=0

We get rid of parentheses

-2x^2+2x-3x-6+9=0

We add all the numbers together, and all the variables

-2x^2-1x+3=0

a = -2; b = -1; c = +3;
Δ = b2-4ac
Δ = -12-4·(-2)·3
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-5}{2*-2}=\frac{-4}{-4}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+5}{2*-2}=\frac{6}{-4}
=-1+1/2
$