In the quadratic equation x^2-3t=(v/3)x, t and v are constants. what are the solutions for x?

let's first off, recall that the factoring of the quadratic, will come from the factoring of the constant, whose FOIL factors added will give the middle term.

now, let's firstly multiply both sides by the LCD of 3, to do away with the denominator.

now, since our constant of -9, can only be factored as 3*3 OR 9*1, so

now, we could have also used the rational root test for that, and get the same variations.

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