Second degree equation solutions involving and / Or

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Lets suppose we have this eq: $x^2-x -2 =0$. It factors into $(x+1)(x-2)=0$, if we solve it as an second equation with Delta equals 9, we have to conclude that the equation accepts two solutions: $x_1 = -1$ and $x_2 = 2$. If we solve it as a factored polynomial, we have to conclude that the equation accepts either $ x = -1$ Or $x =2$ and these are very different conclusions if not contradictory. How should I see this?

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Whether you solved your equation with factoring or using the quadratic formula, the equation has two solutions.

The solutions are, as you have mentioned, $x=-1$ and $x=2.$

Of course these solutions are different and there is room for confusion when we use the same notation $x$ for two different values.

In order to avoid confusion, we may use $x_1=-1$ and $x_2=2.$

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As everyone here is saying, but in other words maybe,

The statement "the solutions of the equation are $-1$ and $2$" is correct,

And the statement "$(x-2)(x+1)=0 \implies x=2$ or $x=-1$"is also correct. (Note that the last statement mentioned is same as saying "$x^2-x-2=0 \implies x=2$ or $x=-1$").

So the form of the equation doesn't actually matter, it just about using the words in a proper manner.

I guess (not sure if I am right) that what dazzeled you is the idea that some teachers tell students to use the word "or" when there is the equal sign $=$ and the word "and" when there is the sign $\ne$.

If so, usually teachers mean by that that if you have a fraction having $(x-2)(x+1)$ as its denominator then when asked for the domain of definition of this fraction you must say that we want $x \ne 2$ and $x \ne -1$ because both numbers give me a zero in my denominator and that is unaccepted.(i.e. we want $x$ to be different from both numbers at the same time).

So in the second statement above we can't have $x=2$ and $x=-1$ at the same time (because $x$ takes one value at a time) but here $x$ can be different than $2$ and $-1$ at the same time and that is what we want here.