How do you solve ${(5 x + 1)}^{2} = 25$ $(5x+1)^2=25$ ?

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Answer 1 · 2016-11-07T00:35:15

$x = \frac{4}{5}$ $x = 4/5$ and $x = - \frac{6}{5}$ $x = -6/5$

$\sqrt{{(5 x + 1)}^{2}} = \sqrt{25}$ $sqrt((5x + 1)^2) = sqrt(25)$

$:.5x+1=+-5$

$5x + 1 = 5$ and $5x + 1 = -5$

$5x + 1 - 1 = 5 - 1$

$5x = 4$

$(5x)/5 = 4/5$

$x = 4/5$

and

$5x + 1 - 1 = -5 - 1$

$5x = -6$

$(5x)/5 = -6/5$

$x = -6/5$

How do you solve (5x+1)^2=25(5x+1)2=25(5x+1)^2=25?