How do you solve \[{x^2} - 6x + 9 = 0\] using the quadratic formula?

Hint: A quadratic equation is a polynomial of degree two and the values of the unknown variable for which the value of the function comes out to be zero are called the zeros/factors/solutions of the polynomial equation. Factorization, completing the square, graphs, quadratic formula, etc. are some methods for finding the roots of a quadratic equation. Now, in the question, we have to solve the quadratic equation given in the question by using the quadratic formula, so for that, we will rewrite the equation in the standard equation form and compare the given equation with it, then we will plug in the values of the coefficients in the quadratic formula to get the correct answer.

Complete step by step answer:
The equation given is ${x^2} - 6x + 9 = 0$
On comparing the given equation with the standard quadratic equation $a{x^2} + bx + c = 0$, we get
$a = 1,\,b = - 6,\,c = 9$
The Quadratic formula is given as –
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Putting the values in the above equation, we get –
$
\Rightarrow x = \dfrac{{ - ( - 6) \pm \sqrt {{{( - 6)}^2} - 4(1)(9)} }}{{2(1)}} \\
\Rightarrow x = \dfrac{{6 \pm \sqrt {36 - 36} }}{2} \\
\Rightarrow x = \dfrac{6}{2} \\
   \Rightarrow x = 3 \\
 $
Hence the zeros of the given equation are equal, that is, they both are equal to 3.

Note: An algebraic expression is defined as an expression containing numerical values along with alphabets; when the alphabet representing an unknown variable quantity is raised to some non-negative integer as a power, a polynomial equation is obtained. Usually, when we fail to find the factors of the equation, we use the quadratic formula but in this question, the factors can be made. So it can also be solved by factoring or by using the formula ${a^2} + {b^2} - 2ab = {(a - b)^2}$.