How do you solve \[2{{x}^{2}}-32=0\]?

Hint: We have to solve the problem using a quadratic formula. First we have to find the coefficients of the variables and then substitute them in the quadratic formula. Next we have to simplify it to get the desired results. Here we will consider the middle term as zero because it is not present in the given equation.

Complete step by step answer:
Given equation is
\[2{{x}^{2}}-32=0\]
The quadratic formula for the standard equation is \[a{{x}^{2}}-bx+c=0\] is
\[y=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
 So using our equation we have to find \[a,b,c\] values.
We will get values as
 \[\begin{align}
  & a=2 \\
 & b=0 \\
 & c=-32 \\
\end{align}\]
Now we have to substitute them in the formula we have
\[\Rightarrow \dfrac{-0\pm \sqrt{{{0}^{2}}-4\times 2\times -32}}{2\times 2}\]
First we have to solve the part present inside the square root.
We will get
\[\Rightarrow \dfrac{0\pm \sqrt{0+8\times 32}}{4}\]
\[\Rightarrow \dfrac{0\pm \sqrt{256}}{4}\]
Now we have to find the square root of \[256\].we will get
\[\Rightarrow \dfrac{0\pm 16}{4}\]
Now we have to split it into two terms.
We will get
\[\Rightarrow \dfrac{0+16}{4}\]
\[\Rightarrow \dfrac{0-16}{4}\]
Now we have to do simple addition and subtraction to find the values.
We will get
\[\Rightarrow 4\]
\[\Rightarrow -4\]
So the \[x\]values we will get are
\[x=4,x=-4\].

Note:
We can solve this in another way also. We can transfer the constant term to the RHS side and simplify it. Then we have to apply a square root on both sides of the equation then we will get the required value for x. This is the simplest way to solve the equation. We can use any of the two methods that are discussed above. We cannot use factorization methods because we do not have a middle term to split. It is better to use a quadratic formula to solve this type of question.