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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-1x^2+9x=0

a = -1; b = 9; c = 0;
Δ = b2-4ac
Δ = 92-4·(-1)·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-9}{2*-1}=\frac{-18}{-2}
=+9
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+9}{2*-1}=\frac{0}{-2}
=0
$