9x^2-1=(3x+1)(5x-8)

Solution for 9x^2-1=(3x+1)(5x-8) equation:

9x^2-1=(3x+1)(5x-8)

We move all terms to the left:

9x^2-1-((3x+1)(5x-8))=0

We multiply parentheses ..

9x^2-((+15x^2-24x+5x-8))-1=0


We calculate terms in parentheses: -((+15x^2-24x+5x-8)), so:

(+15x^2-24x+5x-8)

We get rid of parentheses

15x^2-24x+5x-8

We add all the numbers together, and all the variables

15x^2-19x-8

Back to the equation:

-(15x^2-19x-8)


We get rid of parentheses

9x^2-15x^2+19x+8-1=0

We add all the numbers together, and all the variables

-6x^2+19x+7=0

a = -6; b = 19; c = +7;
Δ = b2-4ac
Δ = 192-4·(-6)·7
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-23}{2*-6}=\frac{-42}{-12}
=3+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+23}{2*-6}=\frac{4}{-12}
=-1/3
$