(9x^2+6x+1)=(3x+1)(x-2)

Solution for (9x^2+6x+1)=(3x+1)(x-2) equation:

(9x^2+6x+1)=(3x+1)(x-2)

We move all terms to the left:

(9x^2+6x+1)-((3x+1)(x-2))=0

We get rid of parentheses

9x^2+6x-((3x+1)(x-2))+1=0

We multiply parentheses ..

9x^2-((+3x^2-6x+x-2))+6x+1=0


We calculate terms in parentheses: -((+3x^2-6x+x-2)), so:

(+3x^2-6x+x-2)

We get rid of parentheses

3x^2-6x+x-2

We add all the numbers together, and all the variables

3x^2-5x-2

Back to the equation:

-(3x^2-5x-2)


We add all the numbers together, and all the variables

9x^2+6x-(3x^2-5x-2)+1=0

We get rid of parentheses

9x^2-3x^2+6x+5x+2+1=0

We add all the numbers together, and all the variables

6x^2+11x+3=0

a = 6; b = 11; c = +3;
Δ = b2-4ac
Δ = 112-4·6·3
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-7}{2*6}=\frac{-18}{12}
=-1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+7}{2*6}=\frac{-4}{12}
=-1/3
$