# How do you solve 5=sqrt(r-3)?

Jun 7, 2017

See a solution process below:

#### Explanation:

First, square both sides of the equation to eliminate the radical while keeping the equation balanced:

${5}^{2} = {\left(\sqrt{r - 3}\right)}^{2}$

$25 = r - 3$

Now, add $\textcolor{red}{3}$ to each side of the equation to solve for $r$ while keeping the equation balanced:

$25 + \textcolor{red}{3} = r - 3 + \textcolor{red}{3}$

$28 = r - 0$

$28 = r$

$r = 28$