How do you solve $x+\dfrac{1}{x}=7$?

Hint: In this question we have the expression in the form of the addition of a linear term and a fraction. We will first take the lowest common multiple of the terms in the left-hand side and then use cross multiplication to rearrange the terms and simplify the expression by using the quadratic formula which is $({{x}_{1}},{{x}_{2}})=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, Where $({{x}_{1}},{{x}_{2}})$ are the roots of the equation and $a,b,c$ are the coefficients of the quadratic equation.

Complete step by step solution:
We have the expression as:
$\Rightarrow x+\dfrac{1}{x}=7$
On taking the lowest common multiple in the left-hand side of the expression, we get:
$\Rightarrow \dfrac{{{x}^{2}}+1}{x}=7$
On transferring the term $x$ from the left-hand side to the right-hand side, we get:
$\Rightarrow {{x}^{2}}+1=7x$
On transferring the term $7x$ from the right-hand side to the left-hand side, we get:
$\Rightarrow {{x}^{2}}-7x+1=0$
Now the above expression is in the form of a quadratic equation which has the coefficients as:
In this question we have:
$a=1$
$b=-7$
$c=1$
On substituting the values in the formula, we get:
$\Rightarrow ({{x}_{1}},{{x}_{2}})=\dfrac{-7\pm \sqrt{{{\left( -7 \right)}^{2}}-4(1)(1)}}{2(1)}$
On simplifying the root, we get:
$\Rightarrow ({{x}_{1}},{{x}_{2}})=\dfrac{-7\pm \sqrt{49-4}}{2(1)}$
On simplifying the denominator, we get:
$\Rightarrow ({{x}_{1}},{{x}_{2}})=\dfrac{-7\pm \sqrt{49-4}}{2}$
On simplifying the terms in the square root, we get:
\[\Rightarrow ({{x}_{1}},{{x}_{2}})=\dfrac{-7\pm \sqrt{45}}{2}\]
Now we can write the term $45$ as a product of $5\times 3\times 3$, on substituting, we get:
\[\Rightarrow ({{x}_{1}},{{x}_{2}})=\dfrac{-7\pm \sqrt{5\times 3\times 3}}{2}\]
On taking the term $3$ out of the square root, we get:
\[\Rightarrow ({{x}_{1}},{{x}_{2}})=\dfrac{-7\pm 3\sqrt{5}}{2}\]
Therefore, the value is \[{{x}_{1}}=\dfrac{-7+3\sqrt{5}}{2}\] and \[{{x}_{2}}=\dfrac{-7-3\sqrt{5}}{2}\], which is the required solution.

Note: It is to be remembered that while cross multiplying when a term which is multiplication and division will change into division and multiplication respectively. Similarly, when a term which is positive or negative when transferred across the $=$ sign will become negative or positive respectively. In this question we have the roots in the real form, if the value of the determinant is lesser than zero, then the quadratic equation is called to have complex roots.