How do you solve the quadratic equation ${x^2} + 3x - 1 = 0$ by completing the square?

Hint: To solve the quadratic equation by completing the square we have to transform the equation in ${(x + a)^2} = {b^2}$ . As there is $3$ in the coefficient of $x$ we can write $3\;x$ as $2 \cdot \dfrac{3}{2}x$ and we will add $\dfrac{9}{4}$ both sides of the equation. Then we will get the required form. Then we will square root both sides and get the solution.

Complete step by step answer:
We have given;
${x^2} + 3x - 1 = 0$
We will try to get the whole square form on both sides.
We know that ${(x + a)^2} = {x^2} + 2ax + {a^2}$ . We will try to convert the part ${x^2} + 3x$ in ${(x + a)^2}$ . We can see the coefficient of $x$ is $3$ . We can write $3\;x$ as $2 \cdot \dfrac{3}{2}x$ . Then we have;
$\Rightarrow {x^2} + 2 \cdot \dfrac{3}{2}x - 1 = 0$
To transform this equation in the square form we will add ${\left( {\dfrac{3}{2}} \right)^2}$ that is $\dfrac{9}{4}$ in both side of the equation and get;
$\Rightarrow {x^2} + 2 \cdot \dfrac{3}{2}x + \dfrac{9}{4} - 1 = \dfrac{9}{4}$
Adding $1$ in both sides we get;
$\Rightarrow {x^2} + 2 \cdot \dfrac{3}{2}x + \dfrac{9}{4} = \dfrac{9}{4} + 1$
Simplifying the above equation we get;
$\Rightarrow {x^2} + 2 \cdot \dfrac{3}{2}x + \dfrac{9}{4} = \dfrac{{13}}{4}$
The left side is equal to nothing but ${\left( {x + \dfrac{3}{2}} \right)^2}$ .So we can write;
$\Rightarrow {\left( {x + \dfrac{3}{2}} \right)^2} = \dfrac{{13}}{4}$
Now $\dfrac{{13}}{4}$ is the square of $\pm \dfrac{{\sqrt {13} }}{2}$ .
Removing square from both side of the above equation we get;
$\Rightarrow x + \dfrac{3}{2} = \pm \dfrac{{\sqrt {13} }}{2}$
Subtracting $\dfrac{3}{2}$ from both side we get;
$\Rightarrow x = \pm \dfrac{{\sqrt {13} }}{2} - \dfrac{3}{2}$
So the solution of the given equation is;
$x = \dfrac{{\sqrt {13} - 3}}{2}$
And $x = \dfrac{{ - \sqrt {13} - 3}}{2}$ .

Additional Information:
If there is some coefficient of ${x^2}$ rather than $1$ to convert the equation in the square form we have to write the coefficient of $x$ as multiplication of $2$ , the square root of the coefficient of ${x^2}$and another constant. Then we have to add the square of that extra constant on both sides of the equation.

Note: This is required to convert both sides of the equation in whole square form to solve this kind of question. To convert the equation in whole square form always we have to look at the coefficient of $x$ and we have to try to convert the coefficient in twice of some number and then we have to add both sides of the square of the coefficient.