How do you solve ${x^2} - 8x + 5 = 0$?

Hint: Here we use the standard quadratic equation and will find the roots of the equation comparing the given equation with the standard quadratic equations $a{x^2} + bx + c = 0$ where roots will be defined as \[x = \dfrac{{ - b \pm \sqrt \Delta }}{{2a}}\]

Complete step-by-step answer:
Take the given equation –
${x^2} - 8x + 5 = 0$
Compare the above equation with the standard equation : $a{x^2} + bx + c = 0$
$
   \Rightarrow a = 1 \\
   \Rightarrow b = - 8 \\
   \Rightarrow c = 5 \;
 $
Also, $\Delta = {b^2} - 4ac$
Place the values from the given comparison
$ \Rightarrow \Delta = {( - 8)^2} - 4(1)(5)$
Simplify the above equation –
$ \Rightarrow \Delta = 64 - 20$
Do subtraction –
$ \Rightarrow \Delta = 44$
Take square root on both the sides of the equation –
$ \Rightarrow \sqrt \Delta = \sqrt {44} $
The above equation can be re-written as –
$ \Rightarrow \sqrt \Delta = \sqrt {4 \times 11} $
Simplify the above equation applying the square of the known number.
$ \Rightarrow \sqrt \Delta = 2\sqrt {11} $
Now, roots of the given equation can be expressed as –
 \[x = \dfrac{{ - b \pm \sqrt \Delta }}{{2a}}\]
Place values in the above equation –
 \[x = \dfrac{{ - ( - 8) \pm 2\sqrt {11} }}{2}\]
Product of minus and minus is plus. Simplify the above equation –
 \[x = \dfrac{{8 \pm 2\sqrt {11} }}{2}\]
Take out common from the numerator from both the terms
 \[x = \dfrac{{2(4 \pm \sqrt {11} )}}{2}\]
Common multiple from the numerator and the denominator cancel each other. Therefore remove from the numerator and the denominator.
 \[ \Rightarrow x = 4 \pm \sqrt {11} \]
Therefore \[x = 4 + \sqrt {11} \] or \[x = 4 - \sqrt {11} \]
This is the required solution.
So, the correct answer is “ \[x = 4 + \sqrt {11} \] or \[x = 4 - \sqrt {11} \] ”.

Note: Be careful regarding the sign convention. Always remember that the square of negative number or the positive number is always positive. Also, product of two negative numbers is always positive whereas, product of one positive and one negative number gives us the negative number.
Every quadratic polynomial has almost two roots. Remember, the quadratic equations having coefficients as the rational numbers has the irrational roots. Also, the quadratic equations whose coefficients are all the distinct irrationals but both the roots are the rational.