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

Solution for 6+(x+3)(x-3)=2x+x equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We use the square of the difference formula

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2-3x-3=0

a = 1; b = -3; c = -3;
Δ = b2-4ac
Δ = -32-4·1·(-3)
Δ = 21
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{-(-3)-\sqrt{21}}{2*1}=\frac{3-\sqrt{21}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{21}}{2*1}=\frac{3+\sqrt{21}}{2}
$