How do you solve ${{x}^{2}}-6x+25=0$?

Hint: We have been given a quadratic equation of $x$ as ${{x}^{2}}-6x+25=0$. We use the quadratic formula to solve the value of the $x$. we have the solution in the form of $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ for general equation of $a{{x}^{2}}+bx+c=0$. We put the values and find the solution.

Complete step by step solution:
We know for a general equation of quadratic $a{{x}^{2}}+bx+c=0$, the value of the roots of $x$ will be $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. This is the quadratic equation solving method. The root part $\sqrt{{{b}^{2}}-4ac}$ of $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ is called the discriminant of the equation.
In the given equation we have ${{x}^{2}}-6x+25=0$. The values of a, b, c is $1,-6,25$ respectively.
We put the values and get $x$ as \[x=\dfrac{-\left( -6 \right)\pm \sqrt{{{\left( -6 \right)}^{2}}-4\times 25\times 1}}{2\times 1}=\dfrac{6\pm \sqrt{-64}}{2}=\dfrac{6\pm 8i}{2}=3\pm 4i\]
The roots of the equation are imaginary numbers. So, values of x are $x=3\pm 4i$.
The discriminant value being negative square, we get the imaginary numbers root values.
In this case the value of $D=\sqrt{{{b}^{2}}-4ac}$ is non-square. ${{b}^{2}}-4ac={{\left( -6 \right)}^{2}}-4\times 25\times 1=-64$.
This is a negative square value. That’s why the roots are imaginary.

Note: We have been given the equation ${{x}^{2}}-2x-4=0$. We form the square part in ${{x}^{2}}-6x+25$.
The square form of subtraction of two numbers be ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$.
We have ${{x}^{2}}-6x+25={{x}^{2}}-2\times x\times 3+{{3}^{2}}+16$.
Forming the square, we get ${{x}^{2}}-6x+25={{\left( x-3 \right)}^{2}}+{{4}^{2}}$.
We get ${{\left( x-3 \right)}^{2}}+{{4}^{2}}=0$. Taking solution, we get
$\begin{align}
  & {{\left( x-3 \right)}^{2}}+{{4}^{2}}=0 \\
 & \Rightarrow {{\left( x-3 \right)}^{2}}=-{{4}^{2}} \\
 & \Rightarrow \left( x-3 \right)=\pm 4i \\
 & \Rightarrow x=3\pm 4i \\
\end{align}$.
Thus, the solution of the equation ${{x}^{2}}-2x-4=0$ is $x=3\pm 4i$.