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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2+3x=0

a = 1; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·1·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*1}=\frac{-6}{2}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*1}=\frac{0}{2}
=0
$