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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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


We calculate terms in parentheses: -(6(x-3)3x), so:

6(x-3)3x

We multiply parentheses

18x^2-54x

Back to the equation:

-(18x^2-54x)


We get rid of parentheses

-18x^2+3x+54x+6=0

We add all the numbers together, and all the variables

-18x^2+57x+6=0

a = -18; b = 57; c = +6;
Δ = b2-4ac
Δ = 572-4·(-18)·6
Δ = 3681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{3681}=\sqrt{9*409}=\sqrt{9}*\sqrt{409}=3\sqrt{409}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(57)-3\sqrt{409}}{2*-18}=\frac{-57-3\sqrt{409}}{-36}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(57)+3\sqrt{409}}{2*-18}=\frac{-57+3\sqrt{409}}{-36}
$