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

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To solve this, we first need to distribute:

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

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

Now we can substitute these back into the original equation:

(x^2 - 2x - 15) + (x^2 - 9) = 0
2x^2 - 2x - 15 - 9 = 0
2x^2 - 2x - 24 = 0

We can simplify further by dividing by 2:

x^2 - x - 12 = 0

Now we need to factor this quadratic equation:

(x-4)(x+3) = 0

So the possible solutions are x = 4 or x = -3.