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+13x-5

Back to the equation:

-(6x^2+13x-5)


We get rid of parentheses

9x^2-6x^2-13x+5-1=0

We add all the numbers together, and all the variables

3x^2-13x+4=0

a = 3; b = -13; c = +4;
Δ = b2-4ac
Δ = -132-4·3·4
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-11}{2*3}=\frac{2}{6}
=1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+11}{2*3}=\frac{24}{6}
=4
$