How do you solve $- 3 {(x + 5)}^{2} + 17 = 2$ $-3(x+5)^2+17=2$ ?

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How do you solve $- 3 {(x + 5)}^{2} + 17 = 2$ $-3(x+5)^2+17=2$ ?

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Answer 1 · 2017-06-16T03:10:48

$x = - 5 \pm \sqrt{5}$ $x = - 5 +- sqrt5$

Explanation:

$- 3 {(x + 5)}^{2} + 17 = 2$ $- 3(x + 5)^2 + 17 = 2$
$- 3 {(x + 5)}^{2} = - 15$ $- 3(x + 5)^2 = - 15$
${(x + 5)}^{2} = 5$ $(x + 5)^2 = 5$
$x + 5 = \pm \sqrt{5}$ $x + 5 = +- sqrt5$
$x = - 5 \pm \sqrt{5}$ $x = - 5 +- sqrt5$

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How do you solve $- 3 {(x + 5)}^{2} + 17 = 2$ $-3(x+5)^2+17=2$ ?

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How do you solve $- 3 {(x + 5)}^{2} + 17 = 2$ $-3(x+5)^2+17=2$ ?