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**Hint:**We first try to take common terms out from the given equation $2{{x}^{2}}+8x=0$. We need to form factorisation from the left side equation $2{{x}^{2}}+8x$. We first divide the equation with 2 as that can be taken as common. Then we have only the variable $x$ to take as common. From the multiplication we find the solution for $2{{x}^{2}}+8x=0$.

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

We need to find the solution of the given equation $2{{x}^{2}}+8x=0$.

We divide the equation with 2 as that can be taken as common.

So, $2{{x}^{2}}+8x=2\left( {{x}^{2}}+4x \right)$. This gives ${{x}^{2}}+4x=0$.

First, we try to take a common number or variable out of the terms ${{x}^{2}}$ and $4x$.

The only thing that can be taken out is $x$.

So, ${{x}^{2}}+4x=x\left( x+4 \right)=0$.

The multiplication of two terms gives 0. This gives that at least one of the terms has to be zero.

We get the values of x as either $x=0$ or $\left( x+4 \right)=0$.

This gives $x=0,-4$.

**The given quadratic equation has 2 solutions and they are $x=0,-4$.**

**Note:**The highest power of the variable or the degree of a polynomial decides the number of roots or the solution of that polynomial. Quadratic equations have 2 roots. Cubic polynomials have 3. It can be both real and imaginary roots.

We can also apply the quadratic equation formula to solve the equation $2{{x}^{2}}+8x=0$.

We know for a general equation of quadratic $a{{x}^{2}}+bx+c=0$, the value of the roots of x will be $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.

In the given equation we have $2{{x}^{2}}+8x=0$. The values of a, b, c are $2,8,0$ respectively.

We put the values and get x as $x=\dfrac{-8\pm \sqrt{{{8}^{2}}-4\times 2\times 0}}{2\times 2}=\dfrac{-8\pm \sqrt{{{8}^{2}}}}{4}=\dfrac{-8\pm 8}{4}=0,-4$.

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