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

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

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

We move all terms to the left:

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

We multiply parentheses ..

x^2-(-(+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

-1x^2+8x-15

Back to the equation:

-(-1x^2+8x-15)


We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2-8x+6=0

a = 2; b = -8; c = +6;
Δ = b2-4ac
Δ = -82-4·2·6
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4}{2*2}=\frac{4}{4}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4}{2*2}=\frac{12}{4}
=3
$