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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((+3x^2-9x+x-3)), so:

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

We get rid of parentheses

3x^2-9x+x-3

We add all the numbers together, and all the variables

3x^2-8x-3

Back to the equation:

-(3x^2-8x-3)


We add all the numbers together, and all the variables

9x^2+6x-(3x^2-8x-3)+1=0

We get rid of parentheses

9x^2-3x^2+6x+8x+3+1=0

We add all the numbers together, and all the variables

6x^2+14x+4=0

a = 6; b = 14; c = +4;
Δ = b2-4ac
Δ = 142-4·6·4
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-10}{2*6}=\frac{-24}{12}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+10}{2*6}=\frac{-4}{12}
=-1/3
$