How do you solve $2{{x}^{2}}+5x+2=0$ ?

Hint: This is a quadratic equation in the form of $a{{x}^{2}}+bx+c=0$ . In this question, we have to solve the equation in order to find the roots. First we need to split the middle term i.e. 5x and then find the factors. After that equating those factors with 0 to find the roots. This is the easiest approach to this question.

Complete step by step answer:
Now, let’s begin to solve the question.
First, write the equation as given in question:
$\Rightarrow 2{{x}^{2}}+5x+2=0$
Now, split the middle term in such a way that we can produce one common factor.
$\begin{align}
  & \Rightarrow 2{{x}^{2}}+4x+x+2=0 \\
 & \Rightarrow 2x(x+2)+(x+2)=0 \\
\end{align}$
See, here ( x + 2 ) is a common factor. Now, we will write this factor only once.
$\Rightarrow \left( 2x+1 \right)(x+2)=0$
Now, equate both the factors with 0 to find the roots.
$\begin{align}
  & \Rightarrow 2x+1=0,\text{ }x+2=0 \\
 & \therefore x=\dfrac{-1}{2},\text{ }x=-2 \\
\end{align}$
So, finally we got the roots of the quadratic equation.

Note: There is an alternative method to find the roots of the quadratic equation. First write the equation:
$\Rightarrow 2{{x}^{2}}+5x+2=0$
As we know the quadratic form $a{{x}^{2}}+bx+c=0$.
So, a = 2, b = 5, c = 2
Formula to find the roots is:
$\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Now, place the values of a, b and c in the above formula.
$\begin{align}
  & \Rightarrow x=\dfrac{-5\pm \sqrt{{{\left( -5 \right)}^{2}}-4\left( 2 \right)\left( 2 \right)}}{2\left( 2 \right)} \\
 & \Rightarrow x=\dfrac{-5\pm \sqrt{25-16}}{4} \\
 & \Rightarrow x=\dfrac{-5\pm \sqrt{9}}{4} \\
 & \Rightarrow x=\dfrac{-5\pm 3}{4} \\
 & \Rightarrow x=\dfrac{-5+3}{4},\text{ }x=\dfrac{-5-3}{4} \\
 & \Rightarrow x=\dfrac{-2}{4},\text{ }x=\dfrac{-8}{4} \\
\end{align}$
After cancellation,
$\therefore x=\dfrac{-1}{2},\text{ }x=-2$
So, you can see we got the same result.
Both the methods are easy and valid. It’s up to you which one to choose. The mistake can be done while splitting the middle term. Middle term splitting is done in such a way so that we can extract common factors from the whole equation.