How do you solve ${(x - 2)}^{2} = 25$ $(x-2)^2=25$ ?

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Answer 1 · 2016-09-12T16:48:06

$x = - 3, 7$ $x = - 3, 7$

We have: ${(x - 2)}^{2} = 25$ $(x - 2)^(2) = 25$

First, let's take the square root of both sides of the equation:

$\Rightarrow x - 2 = \pm 5$ $=> x - 2 = pm 5$

Then, let's add $2$ $2$ to both sides:

$\Rightarrow x = 2 \pm 5$ $=> x = 2 pm 5$

$\Rightarrow x = - 3, 7$ $=> x = - 3, 7$

Therefore, the solutions to the equation are $x = - 3$ $x = - 3$ and $x = 7$ $x = 7$ .

How do you solve (x-2)^2=25(x−2)2=25(x-2)^2=25?