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

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

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

We move all terms to the left:

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


We calculate terms in parentheses: -(3x(3x+9)), so:

3x(3x+9)

We multiply parentheses

9x^2+27x

Back to the equation:

-(9x^2+27x)


We get rid of parentheses

-9x^2+6x-27x+18=0

We add all the numbers together, and all the variables

-9x^2-21x+18=0

a = -9; b = -21; c = +18;
Δ = b2-4ac
Δ = -212-4·(-9)·18
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-33}{2*-9}=\frac{-12}{-18}
=2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+33}{2*-9}=\frac{54}{-18}
=-3
$