How do you solve $x^{2} - 2 x + 7 = - 4$ $x^2-2x+7=-4$ using the quadratic formula?

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How do you solve $x^{2} - 2 x + 7 = - 4$ $x^2-2x+7=-4$ using the quadratic formula?

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Answer 1 · 2016-11-11T15:13:51

$x = 1 \pm i \sqrt{10}$ $x=1+-isqrt10$

(There are no real solutions.)

Explanation:

Make one side equal to zero by adding 4 to each side:
$x^{2} - 2 x + 7 = - 4$ $x^2-2x+7=-4$
$x^{2} - 2 x + 11 = 0$ $x^2-2x+11=0$

Quadratic formula:
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$
$x = \frac{2 \pm \sqrt{{(- 2)}^{2} - (4) (1) (11)}}{2 \cdot 1}$ $x=(2+-sqrt[(-2)^2-(4)(1)(11)])/(2*1)$

Simplify:
$x = \frac{2 \pm \sqrt{- 40}}{2}$ $x=(2+-sqrt(-40))/(2)$
$x = \frac{2 \pm 2 i \sqrt{10}}{2}$ $x=(2+-2isqrt10)/2$
$x = 1 \pm i \sqrt{10}$ $x=1+-isqrt10$

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How do you solve $x^{2} - 2 x + 7 = - 4$ $x^2-2x+7=-4$ using the quadratic formula?

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Explanation:

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How do you solve x2−2x+7=−4x^2-2x+7=-4 using the quadratic formula?

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Explanation:

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How do you solve $x^{2} - 2 x + 7 = - 4$ $x^2-2x+7=-4$ using the quadratic formula?