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To add my pinch of salt to the existing answers (this method requires to be quite comfortable with quadratics so not for everyone!)

once you've got to y^2–5y–14=0, you dont necessarily have to fully solve the equation.

Since the product of it's solutions will be -14/1 (c/a formula of a quadratic), we know that one of the y solutions will be negative, one will be positive (because the only way to multiply two numbers to get a negative is you have one negative / one positive)

however y = x^2 so cannot be negative, so the only cases to consider for x are when y>0

for the y>0 solution, there will be two x equivalents one negative, one positive

So for that reason 1 is the answer
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