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**Hint:**From the given question, we have been asked to find the completing square method solution of the question \[2{{x}^{2}}-8x-15=0\].

We can solve the given equation by completing the square method by separating the variable terms in the equation from the constant terms and rearranging the given equation \[2{{x}^{2}}-8x-15=0\].

**Complete step-by-step solution:**

So in the process of solving the equation firstly we have to separate the terms from each other and bring the constant term to the right hand side then the equation will be reduced as follows.

\[\Rightarrow 2{{x}^{2}}-8x=15\]

Here we must make sure that the coefficient of \[{{x}^{2}}\] is always 1 and must not be any other constant so the equation after this will be reduced as follows.

\[\Rightarrow {{x}^{2}}-4x=7.5\]

Here for this equation we will be adding 4 on both the left hand and right hand side of the equation. So the equation will be as follows.

\[\Rightarrow {{x}^{2}}-4x+4=11.5\]

For this equation if we carefully notice the left hand side expression is a complete square so after factoring the expression on the left hand side of the equation we will get the equation reduced to as below.

\[\Rightarrow {{\left( x-2 \right)}^{2}}=11.5\]

For this equation we will take the square root on both the left hand and right hand sides of the equation then the equation will be reduced as follows.

\[\Rightarrow \sqrt{{{\left( x-2 \right)}^{2}}}=\sqrt{11.5}\]

\[\Rightarrow x-2=\pm \sqrt{11.5}\]

**Here now we will bring the 2 on the left hand side to the right hand side so after doing it the equation will be reduced as follows.**

\[\Rightarrow x=2\pm \sqrt{11.5}\]

\[\Rightarrow x=2\pm \sqrt{11.5}\]

**Note:**We should be well aware of the transformations that have done in the question must be done very carefully and must be knowing that here we have two answers as after taking a square root to a constant it will be \[\pm \sqrt{11.5}\] and not just \[\sqrt{11.5}\] so therefore it has two answers which when done by using completing square method.

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