How do you find the roots of $y = x^{2} + 3 x + 6$ $y=x^2+3x+6$ ?

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How do you find the roots of $y = x^{2} + 3 x + 6$ $y=x^2+3x+6$ ?

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Answer 1 · 2014-11-04T15:43:39

This quadratic function has no real roots, but if you are looking for its complex roots, then by Quadratic Formula, we have

$x = \frac{- 3 \pm \sqrt{3^{2} - 4 \cdot 1 \cdot 6}}{2 \cdot 1} = \frac{- 3 \pm \sqrt{- 15}}{2} = \frac{- 3 \pm \sqrt{15} i}{2}$ $x={-3 pm sqrt{3^2-4cdot1cdot6}}/{2cdot1}={-3 pm sqrt{-15}}/2={-3 pm sqrt{15}i}/2$

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How do you find the roots of $y = x^{2} + 3 x + 6$ $y=x^2+3x+6$ ?

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