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

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

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

We move all terms to the left:

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

We use the square of the difference formula

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


We calculate terms in parentheses: -((x+5)), so:

(x+5)

We get rid of parentheses

x+5

Back to the equation:

-(x+5)


We get rid of parentheses

x^2-x-5-9=0

We add all the numbers together, and all the variables

x^2-1x-14=0

a = 1; b = -1; c = -14;
Δ = b2-4ac
Δ = -12-4·1·(-14)
Δ = 57
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{57}}{2*1}=\frac{1-\sqrt{57}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{57}}{2*1}=\frac{1+\sqrt{57}}{2}
$