3x(2x+)=36x+9

Solution for 3x(2x+)=36x+9 equation:

3x(2x+)=36x+9

We move all terms to the left:

3x(2x+)-(36x+9)=0

We add all the numbers together, and all the variables

3x(+2x)-(36x+9)=0

We multiply parentheses

6x^2-(36x+9)=0

We get rid of parentheses

6x^2-36x-9=0

a = 6; b = -36; c = -9;
Δ = b2-4ac
Δ = -362-4·6·(-9)
Δ = 1512
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:

$\sqrt{\Delta}=\sqrt{1512}=\sqrt{36*42}=\sqrt{36}*\sqrt{42}=6\sqrt{42}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-6\sqrt{42}}{2*6}=\frac{36-6\sqrt{42}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+6\sqrt{42}}{2*6}=\frac{36+6\sqrt{42}}{12}
$