How do you solve $x^{2} + 2 x + 13 = - 8$ $x^2 + 2x + 13 = -8$ ?

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How do you solve $x^{2} + 2 x + 13 = - 8$ $x^2 + 2x + 13 = -8$ ?

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Answer 1 · 2017-07-11T16:53:01

See below.

Explanation:

Adding $8$ $8$ to both sides,

$x^{2} + 2 x + 21 = 0$ $x^2 + 2x + 21=0$

Let's use the quadratic formula, as factoring will not work in this case.

$x = \frac{- 2 \pm \sqrt{2^{2} - 4 (1) (21)}}{2} = \frac{- 2 \pm \sqrt{- 80}}{2} = \frac{- 2 \pm 4 \sqrt{- 5}}{2} = \frac{- 2 \pm 4 i \sqrt{5}}{2} = - 1 \pm 2 i \sqrt{5}$ $x=\frac(-2\pm\sqrt(2^2-4(1)(21)))(2)=\frac(-2\pm\sqrt(-80))(2)=\frac(-2\pm4\sqrt(-5))(2)=\frac(-2\pm4i\sqrt(5))(2)=-1\pm2i\sqrt5$

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How do you solve $x^{2} + 2 x + 13 = - 8$ $x^2 + 2x + 13 = -8$ ?

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Science

Math

Humanities

... and beyond

How do you solve x2+2x+13=−8x^2 + 2x + 13 = -8?

1 Answer

Explanation:

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Impact of this question

How do you solve $x^{2} + 2 x + 13 = - 8$ $x^2 + 2x + 13 = -8$ ?