18x(2x-1)=6(x+2)+3x

Solution for 18x(2x-1)=6(x+2)+3x equation:

18x(2x-1)=6(x+2)+3x

We move all terms to the left:

18x(2x-1)-(6(x+2)+3x)=0

We multiply parentheses

36x^2-18x-(6(x+2)+3x)=0


We calculate terms in parentheses: -(6(x+2)+3x), so:

6(x+2)+3x

We add all the numbers together, and all the variables

3x+6(x+2)

We multiply parentheses

3x+6x+12

We add all the numbers together, and all the variables

9x+12

Back to the equation:

-(9x+12)


We get rid of parentheses

36x^2-18x-9x-12=0

We add all the numbers together, and all the variables

36x^2-27x-12=0

a = 36; b = -27; c = -12;
Δ = b2-4ac
Δ = -272-4·36·(-12)
Δ = 2457
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:

$\sqrt{\Delta}=\sqrt{2457}=\sqrt{9*273}=\sqrt{9}*\sqrt{273}=3\sqrt{273}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-3\sqrt{273}}{2*36}=\frac{27-3\sqrt{273}}{72}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+3\sqrt{273}}{2*36}=\frac{27+3\sqrt{273}}{72}
$