How do you solve $\sqrt{2 x - 3} = 5$ $sqrt(2x - 3) = 5$ ?

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How do you solve $\sqrt{2 x - 3} = 5$ $sqrt(2x - 3) = 5$ ?

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Answer 1 · 2018-03-02T09:52:54

We have, $\sqrt{2 x - 3} = 5$ $sqrt(2x-3)=5$
Squaring on both sides,we have,
$2 x - 3 = {(5)}^{2}$ $2x-3=(5)^2$
or, $2 x = 25 + 3$ $2x=25+3$
or, $x = 28 / 2$ $x=28//2$ which is 14.

Thus,x= $14$ $14$

Answer 2 · 2018-03-02T09:53:07

$x = 14$ $x=14$

Explanation:

$\sqrt{2 x - 3} = 5$ $sqrt(2x−3)=5$
$2 x - 3 = 25$ $2x-3=25$
$2 x = 25 + 3$ $2x=25+3$
$x = \frac{28}{2}$ $x=28/2$
$x = 14$ $x=14$

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How do you solve $\sqrt{2 x - 3} = 5$ $sqrt(2x - 3) = 5$ ?

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How do you solve $\sqrt{2 x - 3} = 5$ $sqrt(2x - 3) = 5$ ?