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I am lazy, so I use the discriminant and the quadratic equation to solve these instead of factoring.
b^2 -4ac
(-2)^2 -4(3)(-2)
= 28 bigger than 1 and two real roots
X = -b +/- sqrt(b^2-4ac)/2a
X = -(-2) +/- sqrt[(-2)^2 - 4(3)(-2)]/2(3)
X = [2 +/- sqrt(28)]/6
this is exact, less factoring, but approximately........
X = 1.215
X = -0.5486
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How do you solve x³-4x equals 0 algebraically?
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