How do you solve $2{{x}^{2}}+2x-1=0$ ?

Hint: For this question, it is given a quadratic equation which is in the form of $a{{x}^{2}}+bx+c=0$. For the given equation, we have to find the roots or zeros. There are two roots in a quadratic equation. Roots are those values which satisfy the equation and result in zero after placing the roots. To find the roots, here we will use quadratic formula:
$\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

Complete step by step answer:
Now, let’s solve the question.
The equation having the highest degree as 2 and which is in the form of $a{{x}^{2}}+bx+c=0$ is known as quadratic equation where ‘a’ and ‘b’ are coefficients and c is the constant term. It consists of two roots. Roots are those values which satisfy the equation and result in zero after placing the roots in the equation. There are basically two methods of finding roots and an equation. The methods are middle term splitting and by using quadratic formulas. Here, we will be using a quadratic formula because middle splitting is only possible if we can split the middle term and we are getting something common while grouping them in pairs. Quadratic formula is:
 $\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Now, write the equation given in the question.
$\Rightarrow 2{{x}^{2}}+2x-1=0$
Here, a = 2, b = 2 and c = -1. Place all these values in quadratic formula, we will get:
$\Rightarrow x=\dfrac{-2\pm \sqrt{{{\left( 2 \right)}^{2}}-4\left( 2 \right)\left( -1 \right)}}{2\left( 2 \right)}$
On solving further:
$\Rightarrow x=\dfrac{-2\pm \sqrt{4+8}}{4}$
Solve the under root:
$\Rightarrow x=\dfrac{-2\pm \sqrt{12}}{4}$
As we know that we do not have the square root of 12. So let’s find the LCM and represent it in the root form:
$\begin{align}
  & 2\left| \!{\underline {\,
  12 \,}} \right. \\
 & 2\left| \!{\underline {\,
  6 \,}} \right. \\
 & 3\left| \!{\underline {\,
  3 \,}} \right. \\
 & 1\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}$
LCM of 12: $2\times 2\times 3$
Now the expression:
$\Rightarrow x=\dfrac{-2\pm 2\sqrt{3}}{4}$
Simplify further:
$\Rightarrow x=\dfrac{-1\pm\sqrt{3}}{2}$
Write the roots for x:
$\Rightarrow x=\dfrac{{-1}-\sqrt{3}}{2},\dfrac{{-1}+\sqrt{3}}{2}$
So this is the answer.

Note: This question is not as complex as it looks like. For solving such questions, first go through it by solving it roughly by middle term splitting. If it doesn’t work, then use the quadratic formula. One more suggestion is if you find any such number whose direct square root is not there, you will solve that root by taking LCM and at the end, reduce the terms as much as possible.