(x-3)(x+3)=^-8(x-2)

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

(x-3)(x+3)=^-8(x-2)

We move all terms to the left:

(x-3)(x+3)-(^-8(x-2))=0

We use the square of the difference formula

x^2-(^-8(x-2))-9=0


We calculate terms in parentheses: -(^-8(x-2)), so:

^-8(x-2)

determiningTheFunctionDomain
-8(x-2)+^

We add all the numbers together, and all the variables

-8(x-2)

We multiply parentheses

-8x+16

Back to the equation:

-(-8x+16)


We get rid of parentheses

x^2+8x-16-9=0

We add all the numbers together, and all the variables

x^2+8x-25=0

a = 1; b = 8; c = -25;
Δ = b2-4ac
Δ = 82-4·1·(-25)
Δ = 164
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{41}}{2*1}=\frac{-8-2\sqrt{41}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{41}}{2*1}=\frac{-8+2\sqrt{41}}{2}
$