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**Hint:**We will solve the above question by using the formula \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] . First, we will put the values in the above equation from the given question according to the general quadratic equation. Then, I will find the values of x.

**Complete step by step answer:**

Let us solve the question.

We have given the equation as \[9{{x}^{2}}-16=0\] .

As we know, the general form of the quadratic equation is \[a{{x}^{2}}+bx+c=0\] .

And the values of x according to the equation \[a{{x}^{2}}+bx+c=0\] are \[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] .

According to the above equations, we can say that

a=9, b=0, and c=-16.

Now, we will plug them in the equation \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] , we get

\[x=\dfrac{-0\pm \sqrt{{{0}^{2}}-4\times 9\times (-16)}}{2\times 9}\]

Now, we will simplify the above equation.

\[\Rightarrow x=\dfrac{\pm \sqrt{-4\times 9\times (-16)}}{2\times 9}\]

\[\Rightarrow x=\pm \dfrac{\sqrt{36\times 16}}{18}\]

\[\Rightarrow x=\pm \dfrac{24}{18}\]

\[\Rightarrow x=\pm \dfrac{4}{3}\]

Hence, the values of x are \[\dfrac{4}{3}\] and \[-\dfrac{4}{3}\] .

**Note:**

In this type of question, we should have a proper knowledge in quadratic equations. We should be careful of the signs. We can solve this by a second method simply. Suppose, we have the equation\[9{{x}^{2}}-16=0\]

We will take the number 16 to the right side of the equation. After that, we will find the value of x.

\[\Rightarrow {{x}^{2}}=\dfrac{16}{9}\]

Now, by taking the square root both sides of the above equation, we get

\[\Rightarrow x=\sqrt{\dfrac{16}{9}}\]

\[\Rightarrow x=\pm \dfrac{4}{3}\]

Therefore, we get the values of x as \[\dfrac{4}{3}\] and \[-\dfrac{4}{3}\].

We have another method of solving the question.

We know that \[{{a}^{2}}-{{b}^{2}}=(a+b)(a-b)\] .

Using this formula, we can find the solution.

Therefore, we can say that \[9{{x}^{2}}-16=(3x+4)(3x-4)\] .

Then, (3x+4)(3x-4)=0

Hence, we can say that \[x=-\dfrac{4}{3}\] and \[x=\dfrac{4}{3}\] .

As we can say that all above methods are giving satisfactory solutions. So, we can use any of the methods to solve them.

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