How do you solve $2{{x}^{2}}+3x-2=0$?

Hint: To solve the quadratic equation given above i.e. $2{{x}^{2}}+3x-2=0$, we are going to use the factorization method to factorize quadratic expression i.e. $2{{x}^{2}}+3x-2$ in which we will multiply the coefficient of ${{x}^{2}}$ and the constant and then write the factors of this multiplication. After that, we will add or subtract these factors in such a way so that we will get 3 which is the coefficient of $x$. And then we will take common from the first two terms and also common from last two terms. And hence, we will get the factors. After that, we will equate each factor to 0 to get the values of $x$.

Complete step by step answer:
In the above problem, we are asked to solve the following quadratic equation:
$2{{x}^{2}}+3x-2=0$
Now, we are going to factorize L.H.S of the above equation by multiplying the coefficient of ${{x}^{2}}$ and the constant which is equal to multiplying 2 by 2 which is equal to 4.
Writing factors of 4 we get,
$\begin{align}
  & 4=1\times 4 \\
 & 4=2\times 2 \\
\end{align}$
Now, if we take the first factorization of 4 then we will get the factors 1 and 4 and then subtracting 4 and 1 and writing in place of 3 we get,
$2{{x}^{2}}+\left( 4-1 \right)x-2=0$
Opening the bracket in the above equation we get,
$2{{x}^{2}}-x+4x-2=0$
Taking $x$ as common from first two terms in the above equation and also taking 2 as common from the last two terms from the above equation we get,
$x\left( 2x-1 \right)+2\left( 2x-1 \right)=0$
Now, taking $\left( 2x-1 \right)$ as common from the L.H.S of the above equation we get,
$\left( 2x-1 \right)\left( x+2 \right)=0$
Equating each of the bracket to 0 we get,
$\begin{align}
  & 2x-1=0 \\
 & \Rightarrow x=\dfrac{1}{2} \\
 & x+2=0 \\
 & \Rightarrow x=-2 \\
\end{align}$
Hence, we got the solutions of the above equation as:
$\begin{align}
  & x=\dfrac{1}{2}; \\
 & x=-2 \\
\end{align}$

Note: We can check the solutions of $x$ that we got in the above solution by substituting these values one by one in the given quadratic equation and see whether these values are satisfying the given equation or not.
Substituting the value of $x=\dfrac{1}{2}$ in the above quadratic equation we get,
$\begin{align}
  & 2{{x}^{2}}+3x-2=0 \\
 & \Rightarrow 2{{\left( \dfrac{1}{2} \right)}^{2}}+3\left( \dfrac{1}{2} \right)-2=0 \\
 & \Rightarrow \dfrac{1}{2}+\dfrac{3}{2}-2=0 \\
 & \Rightarrow \dfrac{4}{2}-2=0 \\
 & \Rightarrow 2-2=0 \\
 & \Rightarrow 0=0 \\
\end{align}$
As L.H.S is equal to R.H.S so the value of $x=\dfrac{1}{2}$ that we have solved above is correct.
Now, substituting the value of $x=-2$ in the given equation we get,
$2{{x}^{2}}+3x-2=0$
\[\begin{align}
  & \Rightarrow 2{{\left( -2 \right)}^{2}}+3\left( -2 \right)-2=0 \\
 & \Rightarrow 2\left( 4 \right)-6-2=0 \\
 & \Rightarrow 8-8=0 \\
 & \Rightarrow 0=0 \\
\end{align}\]
As L.H.S is equal to R.H.S so the value of $x=-2$ that we have solved above is correct.