How do you solve $3 {(y - 5)}^{2} = - 12$ $3(y-5)^2=-12$ ?

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How do you solve $3 {(y - 5)}^{2} = - 12$ $3(y-5)^2=-12$ ?

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Answer 1 · 2017-07-08T23:22:20

2 imaginary roots:
$y = 5 \pm 2 i$ $y = 5 +- 2i$

Explanation:

${(y - 5)}^{2} = - \frac{12}{3} = - 4 = 4 i^{2}$ $(y - 5)^2 = - 12/3 = - 4 = 4i^2$
$y - 5 = \pm 2 i$ $y - 5 = +- 2i$
There are 2 imaginary roots:
$y = 5 \pm 2 i$ $y = 5 +- 2i$

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How do you solve $3 {(y - 5)}^{2} = - 12$ $3(y-5)^2=-12$ ?

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How do you solve $3 {(y - 5)}^{2} = - 12$ $3(y-5)^2=-12$ ?