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

Solution for (3x-9)(3x-9)=(2x+2)(2x+29) equation:

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

We move all terms to the left:

(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 ..

(+4x^2+58x+4x+58)

We get rid of parentheses

4x^2+58x+4x+58

We add all the numbers together, and all the variables

4x^2+62x+58

Back to the equation:

-(4x^2+62x+58)


We get rid of parentheses

9x^2-4x^2-27x-27x-62x+81-58=0

We add all the numbers together, and all the variables

5x^2-116x+23=0

a = 5; b = -116; c = +23;
Δ = b2-4ac
Δ = -1162-4·5·23
Δ = 12996
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12996}=114$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-116)-114}{2*5}=\frac{2}{10}
=1/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-116)+114}{2*5}=\frac{230}{10}
=23
$