How do you solve it by completing the square: ${{x}^{2}}-14x=51$ ?

Hint: At first, we rewrite the middle term of the expression as $2\times 1x\times $ “some numeric value”. The square of this “numeric value” is then added to both sides of the equation. After rearranging the terms and finally by taking square root on both the sides, we get the desired solution.

Complete step by step answer:
Squaring a quadratic expression simply means representing the quadratic equation as the sum of square of a linear expression plus a constant term. Let us consider the general quadratic expression $a{{x}^{2}}+bx+c$ . Squaring it would mean representing it in the form $a{{\left( x+d \right)}^{2}}+e$ .
The given equation is
${{x}^{2}}-14x=51$
At first, we have to bring $51$ to the left hand side of the equation. We get,
$\Rightarrow {{x}^{2}}-14x-51=0$
As we can see that the coefficient of ${{x}^{2}}$ is $1$ . We can express it as ${{\left( 1x \right)}^{2}}$ . This means that the ${{x}^{2}}$ term requires no further modification. So, we now try to focus on the middle term that is $-14x$ and think of some way to express it as of the form $2\times 1x\times $ “some numeric value”. This numeric value can be nothing except $-7$ . The equation thus becomes,
$\Rightarrow {{x}^{2}}+2\times \left( 1x \right)\times \left( -7 \right)-51=0....equation1$
We know that the formula for squaring of a linear expression of the form $ax+b$ is
${{\left( ax+b \right)}^{2}}={{a}^{2}}{{x}^{2}}+2\times \left( ax \right)\times b+{{b}^{2}}....equation2$
In our problem, $a=1$ , and $b=-7$ . Therefore, if we somehow add a ${{\left( -7 \right)}^{2}}$ to $equation1$ , we can show it as a square of a linear term. We do this by adding ${{\left( -7 \right)}^{2}}$ to both sides of $equation1$ . The equation thus becomes,
$\Rightarrow {{x}^{2}}+2\times \left( 1x \right)\times \left( -7 \right)+{{\left( -7 \right)}^{2}}-51={{\left( -7 \right)}^{2}}$
The first three terms can be clubbed together and expressed as a square of a linear expression according to $equation2$ . The equation thus becomes,
$\Rightarrow {{\left( x+\left( -7 \right) \right)}^{2}}-51={{\left( -7 \right)}^{2}}$
Adding $51$ to both sides of the equation, we get,
$\Rightarrow {{\left( x+\left( -7 \right) \right)}^{2}}={{\left( -7 \right)}^{2}}+51$
Now, ${{\left( -7 \right)}^{2}}=49$ . So, the equation gets,
$\Rightarrow {{\left( x-7 \right)}^{2}}=49+51$
Adding $49$ and $51$ we get,
$\Rightarrow {{\left( x-7 \right)}^{2}}=100....equation3$
Now, the given equation has been completely transformed to another form ${{z}^{2}}=g$ where $z$ is the variable and $g$ is a constant. This can be solved by taking square root on both sides. Thus, taking square roots on both sides of $equation3$ we get,
$\Rightarrow \left( x-7 \right)=\pm 10....equation4$
$\pm $ indicates that there are two solutions to $\left( x-7 \right)$ , one positive and the other negative. Adding $7$ to both sides of the equation, we get,
$\Rightarrow x=7\pm 10$
This gives two values of $x$ , one being $7+10=17$ and the other being $7-10=-3$ .

Therefore, we can conclude that the solution of ${{x}^{2}}-14x=51$ by completing the squares are $x=17$ and $x=-3$

Note: We must be careful while transforming the middle term of the quadratic expression as most of the students commit mistakes here. We must always start with the middle term and depending upon the middle term only, another square value is added to make a perfect square. Finally, we must not forget to include both the positive and the negative values at the time of square rooting.