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

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

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

We move all terms to the left:

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

We multiply parentheses ..

-((+x^2-5x+3x-15))+9=0


We calculate terms in parentheses: -((+x^2-5x+3x-15)), so:

(+x^2-5x+3x-15)

We get rid of parentheses

x^2-5x+3x-15

We add all the numbers together, and all the variables

x^2-2x-15

Back to the equation:

-(x^2-2x-15)


We get rid of parentheses

-x^2+2x+15+9=0

We add all the numbers together, and all the variables

-1x^2+2x+24=0

a = -1; b = 2; c = +24;
Δ = b2-4ac
Δ = 22-4·(-1)·24
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-10}{2*-1}=\frac{-12}{-2}
=+6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+10}{2*-1}=\frac{8}{-2}
=-4
$