Question 453570
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Divide both sides of the equation by -3.

Factor *[tex \Large x] out of the resulting binomial in the LHS.

Apply the zero product rule.  Solve each resulting linear equation.  If *[tex \Large x_1] and *[tex \Large x_2] are the solution to the linear equations formed by setting each of the linear factors equal to zero, then the roots (zeros or *[tex \Large x]-intercepts) of the given quadratic are *[tex \Large \left(x_1,\,0\right)] and *[tex \Large \left(x_2,\,0\right)]

John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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