How do you solve ${{x}^{2}}-45=0$ using the quadratic formula?

Hint: We have to use the quadratic formula to solve the given quadratic equation. We will look at the standard quadratic equation. Then we will compare the given quadratic equation with the standard quadratic equation and obtain the values of the corresponding coefficients. We will see the quadratic formula and then substitute the respective values in this formula to obtain the solution.

Complete step by step answer:
The given quadratic equation is ${{x}^{2}}-45=0$. The standard quadratic equation is given as $a{{x}^{2}}+bx+c=0$ where $a$, $b$ and $c$ are the coefficients and $x$ is the variable. Now, we will compare the given quadratic equation with the standard quadratic equation. We get the values of the corresponding coefficients as $a=1$, $b=0$ and $c=-45$.
The quadratic formula is given by
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Now, we will substitute the values $a=1$, $b=0$ and $c=-45$ in the above formula. Substituting the values, we get the following,
$x=\dfrac{-\left( 0 \right)\pm \sqrt{{{\left( 0 \right)}^{2}}-4\left( 1 \right)\left( -45 \right)}}{2\times 1}$
Simplifying the above equation, we get
$x=\pm\dfrac{\sqrt{180}}{2}$
We can write $180=36\times 5$. So, we get the following,
\[\begin{align}
  & x=\pm\dfrac{\sqrt{36\times 5}}{2} \\
 & \Rightarrow x=\pm\dfrac{6\sqrt{5}}{2} \\
 & \therefore x=\pm3\sqrt{5} \\
\end{align}\]

Hence, the solution of the given equation using the quadratic formula is $x=\pm3\sqrt{5}$.

Note: We can check our solution by shifting the constant term to the other side of the equation and then taking the square root on both sides. This will give us the same solution. We can use this to check our solution because the given equation does not have the middle term. There are other methods which can be used to solve quadratic equations. These methods are the completing square method and factorization method.