What is the solution set for $x^{2} + 2 x + 10 = 0$ $x^2 + 2x + 10 = 0$ ?

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What is the solution set for $x^{2} + 2 x + 10 = 0$ $x^2 + 2x + 10 = 0$ ?

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Answer 1 · 2015-08-10T10:24:09

$x = \frac{- 2 + \sqrt{- 36}}{2}$ $x =(-2+ sqrt( - 36))/ (2)$
$x = \frac{- 2 - \sqrt{- 36}}{2}$ $x =(-2- sqrt( - 36))/ (2)$

Explanation:

Since ------- $2^{2}$ $2^2$ - $(4 \times 1 \times 10)$ $(4 xx1 xx 10)$ < 0, x has imaginary roots

$x = \frac{- b \pm \sqrt{b^{2} - (4 a c)}}{2 a}$ $x =(-b+- sqrt(b^2 - (4ac)))/ (2a)$
$x = \frac{- 2 \pm \sqrt{2^{2} - (4 \times 1 \times 10)}}{2 \times 1}$ $x =(-2+- sqrt(2^2 - (4xx1xx10)))/ (2xx1)$
$x = \frac{- 2 \pm \sqrt{4 - 40}}{2}$ $x =(-2+- sqrt(4 - 40))/ (2)$
$x = \frac{- 2 + \sqrt{- 36}}{2}$ $x =(-2+ sqrt( - 36))/ (2)$
$x = \frac{- 2 - \sqrt{- 36}}{2}$ $x =(-2- sqrt( - 36))/ (2)$

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What is the solution set for $x^{2} + 2 x + 10 = 0$ $x^2 + 2x + 10 = 0$ ?

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