How do solve \[{{x}^{2}}+6x=1\]?

Hint: This type of problem is based on the concept of solving quadratic equations. First, we have to subtract 1 from both the sides of the equation and obtain a simplified equation. Then, consider the obtained equation and use quadratic formula to find the value of ‘x’, that is, \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]. Here, a=1, b=6 and c=-1. We need to find the values of the given equation by making necessary calculations. And then solve x by making some adjustments to the given quadratic equation. We get two values of x.

Complete step-by-step answer:
According to the question, we are asked to solve the given equation\[{{x}^{2}}+6x=1\] .
We have been given the equation is \[{{x}^{2}}+6x=1\]. -----(1)
We know that for a quadratic equation \[a{{x}^{2}}+bx+c=0\],
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].
Here, by comparing this with equation (1), we get,
a=1, b=6 and c=-1.
 And also \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].
Therefore,
\[x=\dfrac{-6\pm \sqrt{{{6}^{2}}-4\left( 1 \right)\left( -1 \right)}}{2\left( 1 \right)}\]
\[x=\dfrac{-2\pm \sqrt{36-4\left( -1 \right)}}{2}\]
We know that \[{{6}^{2}}=36\].
Using this in the above obtained expression, we get,
\[x=\dfrac{-2\pm \sqrt{36-4\left( -1 \right)}}{2}\]
On further simplifications, we get,
\[\Rightarrow x=\dfrac{-2\pm \sqrt{4+36}}{2}\]
40 can be written as a product of 4 and 10.
Thus, we get
\[x=\dfrac{-2\pm \sqrt{4\times 10}}{2}\]
We know that \[\sqrt{a\times b}=\sqrt{a}\times \sqrt{b}\].
Using this property, we get,
\[x=\dfrac{-2\pm \sqrt{4}\times \sqrt{10}}{2}\]
On further simplification, we get
\[x=\dfrac{-2\pm 2\times \sqrt{10}}{2}\]
Let us now take 2 common from the obtained expression.
\[\Rightarrow x=\dfrac{2\left( -1\pm \sqrt{10} \right)}{2}\]
Cancelling out 2 from the numerator and denominator, we get
\[x=-1\pm \sqrt{10}\]
Therefore,
The values of ‘x’ are \[-1+\sqrt{10}\] and \[-1-\sqrt{10}\].
Hence, the values of ‘x’ in the quadratic equation \[{{x}^{2}}+6x=1\] are \[-1+\sqrt{10}\] and \[-1-\sqrt{10}\].

Note: Whenever you get this type of problem, we should always try to make the necessary changes in the given equation to get the final solution of the equation which will be the required answer. We should avoid calculation mistakes based on sign conventions. We should always make some necessary calculations to obtain zero in the right-hand side of the equation for easy calculations.