Answer

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**Hint:**In order to solve this question, we will solve the equation as a quadratic equation and we will simplify the equation by finding the roots and by identifying which formula is to use.

**Complete step-by-step solution:**

We have the given equation:

\[{{x}^{2}}-6x-5=0\]

Now for a quadratic equation in the form of $a{{x}^{2}}+bx+c=0$ , we will use the following formula: $\left( {{x}_{1}},{{x}_{2}} \right)=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

Therefore from the above equation and on comparing, we get $a=1,b=-6,c=-5$

Now, substituting the values in formula we get: $\left( {{x}_{1}},{{x}_{2}} \right)=\dfrac{-\left( -6 \right)\pm \sqrt{{{\left( -6 \right)}^{2}}-4\times 1\times \left( -5 \right)}}{2\times 1}$

Now we will first simplify, root part in our equation:

$\Rightarrow \sqrt{36+20}$

$\Rightarrow \sqrt{56}$

Now by doing prime factorization of $56$, we get:

$\Rightarrow \sqrt{{{2}^{3}}\times 7}$

Now by applying exponent rule that is ${{a}^{b+c}}={{a}^{b}}.{{a}^{c}}$ we get:

$\Rightarrow \sqrt{{{2}^{2}}\times 2\times 7}$

Now by applying radical rule that is $\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}$ we get:

$\Rightarrow \sqrt{{{2}^{2}}}\sqrt{2\times 7}$

Now again by applying radical rule that is $\sqrt[n]{{{a}^{n}}}=a$ we get:

$\sqrt{{{2}^{2}}}=2$

$\Rightarrow 2\sqrt{2\times 7}$

Therefore, on simplifying we get:

$\Rightarrow 2\sqrt{14}$

Now coming to our equation and substituting what we equate, we get:

\[\] $\left( {{x}_{1}},{{x}_{2}} \right)=\dfrac{6\pm 2\sqrt{14}}{2}$

Now as we can see that there are common multiple of $2$, so we will use factorization method.

Therefore, on factorization, we get:

$\left( {{x}_{1}},{{x}_{2}} \right)=\dfrac{2\left( 3\pm \sqrt{14} \right)}{2}$

Now, on dividing and equating, we get:

$\left( {{x}_{1}},{{x}_{2}} \right)=3\pm \sqrt{14}$

Now on equating the above equation we get the value of x, i.e.:

${{x}_{1}}=3+\sqrt{14}$ and ${{x}_{2}}=3-\sqrt{14}$

**Therefore, the solution to the given quadratic equation are:**

${{x}_{1}}=3+\sqrt{14}$ and ${{x}_{2}}=3-\sqrt{14}$

${{x}_{1}}=3+\sqrt{14}$ and ${{x}_{2}}=3-\sqrt{14}$

**Note:**There are three forms of quadratic equation, Standard form:$y= a{{x}^{2}}+bx+c$ where $a,b,c$ are just numbers. Factored form:\[~y\text{ }=\text{ }\left( ax\text{ }+\text{ }c \right)\left( bx\text{ }+\text{ }d \right)\] again the $a,b,c,d$ are just numbers. Vertex form: \[y\text{ }=\text{ }a{{\left( x\text{ }+\text{ }b \right)}^{2}}\text{ }+\text{ }c\] again the $a,b,c$ are just numbers.

We have to identify by looking at the equation which form we have to use, then only we can solve the equation.

If we have a standard form of a quadratic equation, we can verify the answer by substituting the answer in the original equation.

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