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

Solution for 9x^2-1=(3x+1)*(2x+5) equation:

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

We move all terms to the left:

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

We multiply parentheses ..

9x^2-((+6x^2+15x+2x+5))-1=0


We calculate terms in parentheses: -((+6x^2+15x+2x+5)), so:

(+6x^2+15x+2x+5)

We get rid of parentheses

6x^2+15x+2x+5

We add all the numbers together, and all the variables

6x^2+17x+5

Back to the equation:

-(6x^2+17x+5)


We get rid of parentheses

9x^2-6x^2-17x-5-1=0

We add all the numbers together, and all the variables

3x^2-17x-6=0

a = 3; b = -17; c = -6;
Δ = b2-4ac
Δ = -172-4·3·(-6)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-19}{2*3}=\frac{-2}{6}
=-1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+19}{2*3}=\frac{36}{6}
=6
$