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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


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

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

5x^2-13x-6

Back to the equation:

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


We get rid of parentheses

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

We add all the numbers together, and all the variables

-4x^2+13x-3=0

a = -4; b = 13; c = -3;
Δ = b2-4ac
Δ = 132-4·(-4)·(-3)
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*-4}=\frac{-24}{-8}
=+3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+11}{2*-4}=\frac{-2}{-8}
=1/4
$