How do you solve ${x^2} + 3x + 21 = 22$ by completing the square $?$

Hint: In this question, we are going to solve the given equation and find the value of $x$.
We are going to solve this equation by using the quadratic formula. The given equation is of the quadratic form.
To find the value of $x$ for the given equation, rewrite the given equation in the quadratic form.
Compare the quadratic form to the given equation and substitute the values in the quadratic formula.

Formula used: The quadratic formula can be written as
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Here ${b^2} - 4ac$is the formula for discriminant.
There are three possible outcomes for the discriminant.
${b^2} - 4ac < 0$, the quadratic equation has no real solutions.
${b^2} - 4ac = 0,$The quadratic equation has only one solution or two real and equal solutions.
${b^2} - 4ac > 0,$The quadratic equation has two real and distinct solutions.

Complete step by step solution:
In this question, we are going to solve the given equation and then find the values of $x$.
First rewrite the given equation in the quadratic form and mark it as $\left( 1 \right)$
${x^2} + 3x + 21 = 22$
Bring the constant term on the right hand side to the left hand side
${x^2} + 3x + 21 - 22 = 0$
${x^2} + 3x - 1 = 0...\left( 1 \right)$
Here $a = 1,\,b = 3,\,c = - 1$
It has a discriminant $\Delta $ given by the formula:
$\Delta = {b^2} - 4ac$
$\Rightarrow$$\Delta = {\left( 3 \right)^2} - 4\left( 1 \right)\left( { - 1} \right)$
$\Rightarrow$$\Delta = 9 + 4$
$\Rightarrow$$\Delta = 13$
Since $\Delta > 0,$ this quadratic equation has two distinct real roots, but since $\Delta $ is not a perfect square, those roots are irrational.
We can find the roots using the quadratic formula.
Substitute those values in the quadratic formula, $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
$ \Rightarrow x = \dfrac{{ - 3 \pm \sqrt {{{\left( 3 \right)}^2} - 4\left( 1 \right)\left( { - 1} \right)} }}{{2\left( 1 \right)}}$
On rewriting we get
$ \Rightarrow x = \dfrac{{ - 3 \pm \sqrt \Delta }}{2}$
Then we get,
$ \Rightarrow x = \dfrac{{ - 3 + \sqrt {13} }}{2},x = \dfrac{{ - 3 - \sqrt {13} }}{2}$
Hence,
$ \Rightarrow x = \dfrac{{ - 3 \pm \sqrt {13} }}{2}$

Thus we get the value of ${x^2} + 3x - 1 = 0$ as $x = \dfrac{{ - 3 \pm \sqrt {13} }}{2}$.

Note: There are various methods to solve the quadratic equation. Some of them are as follows: factoring, using the square roots, completing the square and the quadratic formula.
The discriminant can be positive, zero, or negative, and this determines how many solutions are there to the given quadratic equation.