Solve :
$x^{2} - 8 x + 12 = 0$

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Solution

$x^{2} - 8 x + 12 = 0$
$x^{2} - 6 x - 2 x + 12 = 0$
$x (x - 6) - 2 (x - 6) = 0$
$(x - 6) (x - 2) = 0$
$x = 6, 2$

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Similar questions

∣ ∣ ∣ \frac{x^{2} - 8 x + 12}{x^{2} - 10 x + 21} ∣ ∣ ∣ = - \frac{x^{2} - 8 x + 12}{x^{2} - 10 x + 21}

x \in [2, a) \cup [b, 7)

x^{2} + 8 x + 16 = 0

x^{2} - 8 x + 5 = 0

x + 4 < \sqrt{- x^{2} - 8 x - 12} .

\frac{x^{2} + 12 x + 20}{3 x - 5} = \frac{x^{2} + 8 x + 12}{2 x + 3}

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