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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((+6x^2-9x+2x-3)), so:

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

We get rid of parentheses

6x^2-9x+2x-3

We add all the numbers together, and all the variables

6x^2-7x-3

Back to the equation:

-(6x^2-7x-3)


We get rid of parentheses

9x^2-6x^2+7x+3-1=0

We add all the numbers together, and all the variables

3x^2+7x+2=0

a = 3; b = 7; c = +2;
Δ = b2-4ac
Δ = 72-4·3·2
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-5}{2*3}=\frac{-12}{6}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+5}{2*3}=\frac{-2}{6}
=-1/3
$