6(3x-2)9x=3(3+3x)-18

Solution for 6(3x-2)9x=3(3+3x)-18 equation:

6(3x-2)9x=3(3+3x)-18

We move all terms to the left:

6(3x-2)9x-(3(3+3x)-18)=0

We add all the numbers together, and all the variables

6(3x-2)9x-(3(3x+3)-18)=0

We multiply parentheses

162x^2-108x-(3(3x+3)-18)=0


We calculate terms in parentheses: -(3(3x+3)-18), so:

3(3x+3)-18

We multiply parentheses

9x+9-18

We add all the numbers together, and all the variables

9x-9

Back to the equation:

-(9x-9)


We get rid of parentheses

162x^2-108x-9x+9=0

We add all the numbers together, and all the variables

162x^2-117x+9=0

a = 162; b = -117; c = +9;
Δ = b2-4ac
Δ = -1172-4·162·9
Δ = 7857
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:

$\sqrt{\Delta}=\sqrt{7857}=\sqrt{81*97}=\sqrt{81}*\sqrt{97}=9\sqrt{97}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-117)-9\sqrt{97}}{2*162}=\frac{117-9\sqrt{97}}{324}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-117)+9\sqrt{97}}{2*162}=\frac{117+9\sqrt{97}}{324}
$