How do you solve $3 {(x - 5)}^{2} - 12 = 0$ $3(x-5)^2 - 12 = 0$ ?

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How do you solve $3 {(x - 5)}^{2} - 12 = 0$ $3(x-5)^2 - 12 = 0$ ?

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Answer 1 · 2016-03-17T16:26:21

$x = 5 + 2 = 7 and x = 5 - 2 = 3$ $x=5+2=7 and x= 5-2=3$

Explanation:

$3 {(x - 5)}^{2} = 12$ $3(x-5)^2=12$
$\Rightarrow {(x - 5)}^{2} = \frac{12}{3} = 4$ $=>(x-5)^2=12/3=4$
$\Rightarrow (x - 5) = \pm \sqrt{4} = \pm 2$ $=>(x-5)=+-sqrt4=+-2$
$x = 5 + 2 = 7 and x = 5 - 2 = 3$ $x=5+2=7 and x= 5-2=3$

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How do you solve $3 {(x - 5)}^{2} - 12 = 0$ $3(x-5)^2 - 12 = 0$ ?

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How do you solve 3(x−5)2−12=0 3(x-5)^2 - 12 = 0?

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How do you solve $3 {(x - 5)}^{2} - 12 = 0$ $3(x-5)^2 - 12 = 0$ ?