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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+4x^2+2x+12x+6)-((3x-1)(3x-1))=0


We calculate terms in parentheses: -((3x-1)(3x-1)), so:

(3x-1)(3x-1)

We multiply parentheses ..

(+9x^2-3x-3x+1)

We get rid of parentheses

9x^2-3x-3x+1

We add all the numbers together, and all the variables

9x^2-6x+1

Back to the equation:

-(9x^2-6x+1)


We get rid of parentheses

4x^2-9x^2+2x+12x+6x+6-1=0

We add all the numbers together, and all the variables

-5x^2+20x+5=0

a = -5; b = 20; c = +5;
Δ = b2-4ac
Δ = 202-4·(-5)·5
Δ = 500
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:

$\sqrt{\Delta}=\sqrt{500}=\sqrt{100*5}=\sqrt{100}*\sqrt{5}=10\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-10\sqrt{5}}{2*-5}=\frac{-20-10\sqrt{5}}{-10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+10\sqrt{5}}{2*-5}=\frac{-20+10\sqrt{5}}{-10}
$