How do you solve \[{x^2} + 18x = 10\] by completing the square?

Hint: In this question, we are given a quadratic equation and we have been asked to solve the equation using completing the square. Find the coefficient of $x$ . Find the half of the coefficient, square it and add the resultant number to both the sides of the equation. Now, identify the terms - ${a^2},2ab,{b^2}$ and make the required square. Then, square root both the sides and find the value of $x$ .

Formula used: We have to apply following algebraic formula:
\[{(a + b)^2} = ({a^2} + 2ab + {b^2}).\]

Complete step-by-step solution:
The given expression is as following:
\[{x^2} + 18x = 10\] .
So, if we tally the L.H.S in the above equation with the formula of \[{(a + b)^2}\] , we can say that the term \[2ab\] should be equivalent to the term \[18x\] .
So, if we expand the term \[18x\] , we can rewrite it as \[(2 \times x \times 9)\] .
Therefore, we can say that the term \[a\] and \[b\] in the formula are equivalent to \[x\] and \[9\] .
So, we need to add a term on both sides of the equation that is squared of \[9\] .
So, we have to add \[81\] on both the sides of the given equation.
So, we can rewrite the equation as following:
\[{x^2} + 18x + 81 = 10 + 81.\]
Now, add the constant terms in the R.H.S, we get the following equation:
\[{x^2} + 18x + 81 = 91.\]
Now, if we write the terms in the L.H.S as in the expanded form, we get:
\[{x^2} + 2 \times x \times 9 + {9^2} = 91\] …………...…. (1)
So, by comparing the L.H.S in the above equation with the formula: \[{(a + b)^2} = ({a^2} + 2ab + {b^2})\] , we rewrite L.H.S as following:
\[{x^2} + 2 \times x \times 9 + {9^2} = {(x + 9)^2}\] .
So, we can rewrite the equation \[(1)\] as following:
\[{(x + 9)^2} = 91\] .
Now, if we take the square root on both the sides of the above equation, we can write the following equation:
\[\sqrt {{{(x + 9)}^2}} = \pm \sqrt {91} \] . (We use \[ \pm \] sign as the square root can give us either positive or negative value)
After solving the above equation, we get:
\[(x + 9) = \pm \sqrt {91} \].
Now, take the constant term to the R.H.S from the L.H.S, we get:
\[x = - 9 \pm \sqrt {91} \].

The given equation has two different roots of \[( - 9 + \sqrt {91} )\] and \[( - 9 - \sqrt {91} )\] .

Note: A quadratic equation is the one whose highest degree is $2$ . A quadratic equation can be solved by following two methods also:
Splitting the middle term method
Quadratic formula method
However, you can use splitting the middle term method only when the roots are real. At other times, it is suggested to use the quadratic formula. This formula will directly give you the required values.