Solve:$ - 2{x^2} + 5x - 2 = 0$

Hint: Here, an equation is given, and we are asked to solve it. To solve an equation means to find the value of the unknown variable in that equation. In our given equation, there is an unknown variable $x$ and we need to find the value of it. As it is a quadratic equation, we can use the method of splitting the middle term to solve it. Splitting the middle term is that splitting the coefficient of $x$ into sum or difference of two numbers such that the product of those two numbers must be equal to the product of the constant term and the coefficient of ${x^2}$ and the sum or difference of those numbers should be equal to the coefficient of $x$. We will get two values for $x$ since the given equation is a quadratic equation.

Complete answer:
We are given an equation and we need to solve the equation.
The word solve means to find the value of the variable $x$ in the given equation.
Here the given equation is $ - 2{x^2} + 5x - 2 = 0$
From this we can understand that it is a quadratic equation.
A general form of a quadratic equation is $a{x^2} + bx + c = 0$
Comparing we get $a = - 2,b = 5,c = - 2$
There are many ways to solve a quadratic equation, but we are going to use the splitting the middle term method.
Since the coefficient of ${x^2}$ is more than one, multiply it with the constant term.
The product is $4$
Splitting the middle term is that splitting the coefficient of $x$ into sum or difference of two numbers such that the product of those two numbers must be equal to the product of the constant term and the coefficient of ${x^2}$ and the sum or difference of those numbers should be equal to the coefficient of $x$
Here the coefficient of $x$ is $5$
Let the two numbers be $4$ and $1$
The product of these two numbers is $4$ which is equal to the product of the constant term and the coefficient of ${x^2}$
The sum of the two numbers is $5$ which is equal to the coefficient of $x$
Hence, the given equation becomes
$ - 2{x^2} + 4x + x - 2 = 0$
Now take $ - 2x$ common in the first part and $1$ in the second part of the equation.
$ - 2x\left( {x - 2} \right) + 1\left( {x - 2} \right) = 0$
$\left( {x - 2} \right)\left( {1 - 2x} \right) = 1$
Now equating each part to zero we get the value of $x$
\[x - 2 = 0\]
$x = 2$
$1 - 2x = 0$
$1 = 2x$
$\dfrac{1}{2} = x$
From this we get the solution of the equation to be $2$ and $\dfrac{1}{2}$

Note:
To solve quadratic equations, we can also follow the formula method.
The formula to find solution of the quadratic equation $a{x^2} + bx + c = 0$ is $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Comparing with our given equation we get $a = - 2,b = 5,c = - 2$
$x = \dfrac{{ - 5 \pm \sqrt {{5^2} - 4\left( { - 2} \right)\left( { - 2} \right)} }}{{2\left( { - 2} \right)}}$
We know that ${5^2} = 25$ . Using this we get,
$x = \dfrac{{ - 5 \pm \sqrt {25 - 4\left( 4 \right)} }}{{2\left( { - 2} \right)}}$
$x = \dfrac{{ - 5 \pm \sqrt {25 - 16} }}{{ - 4}}$
$x = \dfrac{{ - 5 \pm \sqrt 9 }}{{ - 4}}$
Same way we know that $\sqrt 9 = 3$
$x = \dfrac{{ - 5 \pm 3}}{{ - 4}}$
From this we get,
$x = \dfrac{{ - 5 + 3}}{{ - 4}},x = \dfrac{{ - 5 - 3}}{{ - 4}}$
$x = \dfrac{{ - 2}}{{ - 4}},x = \dfrac{{ - 8}}{{ - 4}}$
$x = \dfrac{1}{2},x = 2$
This is another way to solve a quadratic equation.