(3x+9)(3x+9)=2x+2(2x+2+27)

Solution for (3x+9)(3x+9)=2x+2(2x+2+27) equation:

(3x+9)(3x+9)=2x+2(2x+2+27)

We move all terms to the left:

(3x+9)(3x+9)-(2x+2(2x+2+27))=0

We add all the numbers together, and all the variables

(3x+9)(3x+9)-(2x+2(2x+29))=0

We multiply parentheses ..

(+9x^2+27x+27x+81)-(2x+2(2x+29))=0


We calculate terms in parentheses: -(2x+2(2x+29)), so:

2x+2(2x+29)

We multiply parentheses

2x+4x+58

We add all the numbers together, and all the variables

6x+58

Back to the equation:

-(6x+58)


We get rid of parentheses

9x^2+27x+27x-6x+81-58=0

We add all the numbers together, and all the variables

9x^2+48x+23=0

a = 9; b = 48; c = +23;
Δ = b2-4ac
Δ = 482-4·9·23
Δ = 1476
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:

$\sqrt{\Delta}=\sqrt{1476}=\sqrt{36*41}=\sqrt{36}*\sqrt{41}=6\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-6\sqrt{41}}{2*9}=\frac{-48-6\sqrt{41}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+6\sqrt{41}}{2*9}=\frac{-48+6\sqrt{41}}{18}
$