How do you solve using the quadratic formula method \[{{x}^{2}}-7x+9=0\]?

Hint: In this problem we have to solve the given quadratic equation and find the value of x using the quadratic formula method. To solve it by using the quadratic formula method, we have to use the quadratic formula given by \[x=\dfrac{-b\pm \sqrt{{{\left( b \right)}^{2}}-4\times a\times \left( c \right)}}{2a}\] to find the value of x. Here, we will substitute the values of a as 1 , b as -7 and c as 9 to get the answer.

Complete step by step solution:
We know that the given quadratic equation is,
\[{{x}^{2}}-7x+9=0\] ….. (1)
We also know that a quadratic equation in standard form is,
\[a{{x}^{2}}+bx+c=0\] ……. (2)
We can now compare the two equations (1) and (2), we get
a = 1, b = -7, c = 9.
We know that the quadratic formula for the standard form \[a{{x}^{2}}+bx+c=0\] is
\[x=\dfrac{-b\pm \sqrt{{{\left( b \right)}^{2}}-4\times a\times \left( c \right)}}{2a}\]
Now we can substitute the value of a, b, c in the above formula, we get
\[\Rightarrow x=\dfrac{-\left( -7 \right)\pm \sqrt{{{\left( -7 \right)}^{2}}-4\times 1\times \left( 9 \right)}}{2\times 1}\]
Now we can simplify the above step, we get
\[\begin{align}
  & \Rightarrow x=\dfrac{\left( 7 \right)\pm \sqrt{49-36}}{2\times 1} \\
 & \Rightarrow x=\dfrac{7\pm \sqrt{13}}{2} \\
\end{align}\]
Now we can separate the terms to simplify it,
\[\begin{align}
  & \Rightarrow x=\dfrac{7}{2}+\dfrac{\sqrt{13}}{2} \\
 & \Rightarrow x=\dfrac{7}{2}-\dfrac{\sqrt{13}}{2} \\
\end{align}\]
Therefore, the value of \[x=\dfrac{7}{2}+\dfrac{\sqrt{13}}{2}\] and \[x=\dfrac{7}{2}-\dfrac{\sqrt{13}}{2}\]

Note: We can also use a simple factorisation method to solve this problem but we have to know that the answer is in root form, so we can solve only by using a quadratic formula to get the final answer correctly. Students may make mistakes in the quadratic formula part by missing out some terms. So, they should be careful and also concentrate while substituting the values in the quadratic formula.