(X+3)(x-3)-2(x+1)=x(-x-1)-3

Solution for (X+3)(x-3)-2(x+1)=x(-x-1)-3 equation:

(X+3)(X-3)-2(X+1)=X(-X-1)-3

We move all terms to the left:

(X+3)(X-3)-2(X+1)-(X(-X-1)-3)=0

We add all the numbers together, and all the variables

(X+3)(X-3)-2(X+1)-(X(-1X-1)-3)=0

We use the square of the difference formula

X^2-2(X+1)-(X(-1X-1)-3)-9=0

We multiply parentheses

X^2-2X-(X(-1X-1)-3)-2-9=0


We calculate terms in parentheses: -(X(-1X-1)-3), so:

X(-1X-1)-3

We multiply parentheses

-1X^2-1X-3

Back to the equation:

-(-1X^2-1X-3)


We add all the numbers together, and all the variables

X^2-(-1X^2-1X-3)-2X-11=0

We get rid of parentheses

X^2+1X^2+1X-2X+3-11=0

We add all the numbers together, and all the variables

2X^2-1X-8=0

a = 2; b = -1; c = -8;
Δ = b2-4ac
Δ = -12-4·2·(-8)
Δ = 65
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{65}}{2*2}=\frac{1-\sqrt{65}}{4}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{65}}{2*2}=\frac{1+\sqrt{65}}{4}
$