How do you solve $ - 4{r^2} + 21r = r + 13$ by completing the square?

Hint: Solving a quadratic equation by completing the square method means finding the value of the variable by changing the equation such that we get a perfect square expression equated to a constant term, i.e. transforming $a{r^2} + br + c = 0$ to ${(r + m)^2} = n$, where $r$ is the only variable. From here we can find the solution by taking square roots on both sides.

Complete step-by-step solution:
In the given equation $ - 4{r^2} + 21r = r + 13$we observe that the degree or the highest power of the variable $r$ is $2$. Therefore, the given equation is a quadratic equation in one variable.
We first shift all the terms to the LHS such that we have only the term $0$ in the RHS.
For this we first subtract $(r + 13)$ from both sides,
$
   - 4{r^2} + 21r - (r + 13) = r + 13 - (r + 13) \\
   \Rightarrow - 4{r^2} + 21r - r - 13 = 0 \\
   \Rightarrow - 4{r^2} + 20r - 13 = 0 \\
 $
Here we got the quadratic equation in its general form. We can observe that $a = - 4$, $b = 20$ and $c = - 13$.
Now we have to transform the equation such that we get a perfect square expression in the LHS and shift the remaining constant term to the RHS {i.e. of the form ${(r + m)^2} = n$}.
We follow the following steps to get a perfect square expression:
Divide both sides of the equation by the coefficient of ${r^2}$, i.e. $a = - 4$.
\[
  \dfrac{{ - 4{r^2} + 20r - 13}}{{ - 4}} = \dfrac{0}{{ - 4}} \\
   \Rightarrow \dfrac{{ - 4}}{{ - 4}}{r^2} + \dfrac{{20}}{{ - 4}}r - \dfrac{{13}}{{ - 4}} = 0 \\
   \Rightarrow {r^2} - 5r + \dfrac{{13}}{4} = 0 \\
 \]
The coefficient of $r$ is $ - 5$, which is equal to $\dfrac{b}{a}$. We can write it as $ - 5 = 2 \times \dfrac{1}{2} \times ( - 5) = 2 \times \dfrac{{ - 5}}{2}$.
\[ \Rightarrow {r^2} + (2 \times r \times \dfrac{{ - 5}}{2}) + \dfrac{{13}}{4} = 0\]
Now we add ${(\dfrac{{ - 5}}{2})^2}$, which is equal to ${(\dfrac{b}{{2a}})^2}$, to both sides of the equation.
\[ \Rightarrow {r^2} + (2 \times r \times \dfrac{{ - 5}}{2}) + {(\dfrac{{ - 5}}{2})^2} + \dfrac{{13}}{4} = 0 + {(\dfrac{{ - 5}}{2})^2}\]
We keep the first three terms on the LHS and shift the constant term $c = \dfrac{{13}}{4}$ to the RHS.
\[ \Rightarrow {r^2} + (2 \times r \times \dfrac{{ - 5}}{2}) + {(\dfrac{{ - 5}}{2})^2} = {(\dfrac{{ - 5}}{2})^2} - \dfrac{{13}}{4}\]
Here we can observe in the LHS that it is forming a perfect square expression of the form \[{r^2} + 2rm + {m^2} = {(r + m)^2}\], where $m = \dfrac{{ - 5}}{2}$.
We simplify the LHS to write the expression in perfect square form,
\[{r^2} + (2 \times r \times \dfrac{{ - 5}}{2}) + {(\dfrac{{ - 5}}{2})^2} = {(r + (\dfrac{{ - 5}}{2}))^2}\]
Or, \[{r^2} + (2 \times r \times \dfrac{{ - 5}}{2}) + {(\dfrac{{ - 5}}{2})^2} = {(r - \dfrac{5}{2})^2}\]
Thus, we transformed the given equation to arrive at a perfect square expression.
Using \[{(r - \dfrac{5}{2})^2}\] in the above equation, we get,
\[
  {(r - \dfrac{5}{2})^2} = {(\dfrac{{ - 5}}{2})^2} - \dfrac{{13}}{4} = \dfrac{{25}}{4} - \dfrac{{13}}{4} \\
   \Rightarrow {(r - \dfrac{5}{2})^2} = \dfrac{{12}}{4} = 3 \\
\]
Now to solve this equation we take square roots on both sides,
\[
   \Rightarrow \sqrt {{{(r - \dfrac{5}{2})}^2}} = \pm \sqrt 3 \\
   \Rightarrow (r - \dfrac{5}{2}) = \pm \sqrt 3 \\
   \Rightarrow r = \dfrac{5}{2} \pm \sqrt 3 \\
\]
We get the value, \[r = \dfrac{5}{2} + \sqrt 3 \] or \[r = \dfrac{5}{2} - \sqrt 3 \].
Thus we solved the given quadratic equation using completing the square method and got the result as \[r = \dfrac{5}{2} + \sqrt 3 \] or \[\dfrac{5}{2} - \sqrt 3 \].


Note: Solving a quadratic equation gives two values as result.
We can summarize the steps to get perfect square expression used in the solution as:
-Divide by $a$.
-Write coefficient of $r$ as $2 \times \dfrac{b}{{2a}}$.
-Add ${(\dfrac{b}{{2a}})^2}$ to both sides of the equation.
-Shift constant term $c$ to the RHS.
-Simplify the LHS to arrive at a perfect square expression.