How do you solve the quadratic equation by completing the square $5{{x}^{2}}+8x-2=0$?

Hint: Now we are given with a quadratic equation in one variable of the form $a{{x}^{2}}+bx+c=0$ first we will divide the whole equation by a. Now we will add and subtract the term ${{\left( \dfrac{b}{2a} \right)}^{2}}$ on both sides. Now we will simplify the equation obtained by using the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ Now we will simplify the equation by taking square root on both sides and solve for x.

Complete step-by-step solution:
Now let us consider the quadratic equation $5{{x}^{2}}+8x-2=0$ .
The given equation is a quadratic equation in the form $a{{x}^{2}}+bx+c=0$ where a = 5 b = 8 and
c = - 2.
Now we want to find the roots of this equation. To do so we will use the completing square method.
Now first we want the coefficient of ${{x}^{2}}$ to be 1.
Hence we will divide the whole equation by 5. Hence we get,
$\Rightarrow {{x}^{2}}+\dfrac{8}{5}x-\dfrac{2}{5}=0$
Now we want to form a complete square on LHS hence we will add and subtract the equation with ${{\left( \dfrac{b}{2a} \right)}^{2}}$ which is nothing but ${{\left( \dfrac{8}{2\times 5} \right)}^{2}}$ .
$\Rightarrow {{x}^{2}}+\dfrac{8}{5}x-\dfrac{2}{5}+{{\left( \dfrac{8}{10} \right)}^{2}}-{{\left( \dfrac{8}{10} \right)}^{2}}=0$
Now we know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ hence using this we get $\begin{align}
  & \Rightarrow {{\left( x+\dfrac{8}{10} \right)}^{2}}-\dfrac{2}{5}-{{\left( \dfrac{8}{10} \right)}^{2}}=0 \\
 & \Rightarrow {{\left( x+\dfrac{8}{10} \right)}^{2}}-\dfrac{2}{5}-\dfrac{64}{100}=0 \\
 & \Rightarrow {{\left( x+\dfrac{8}{10} \right)}^{2}}=\dfrac{64}{100}+\dfrac{2}{5} \\
 & \Rightarrow {{\left( x+\dfrac{8}{10} \right)}^{2}}=\dfrac{64+40}{100} \\
 & \Rightarrow {{\left( x+\dfrac{8}{10} \right)}^{2}}=\dfrac{104}{100} \\
\end{align}$
Now taking square root on both sides we get,
$\Rightarrow x=\dfrac{-8\pm \sqrt{104}}{10}$
Hence the roots of the given equation are $x=\dfrac{-8+\sqrt{104}}{10}$ and $x=\dfrac{-8-\sqrt{104}}{10}$.

Note: Now note that while taking complete square method we do not complete the square using ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ . Also note that while taking square roots in the equation we must take positive and negative cases as ${{\left( -x \right)}^{2}}={{\left( x \right)}^{2}}$ . Hence we will get two solutions for the equation. Also check the solution by substituting it back in the given equation.