How do you solve \[2{x^2}\, + \,5x\, - \,3\, \leqslant \,0\] ?

Hint:Try to solve it using Quadratic factorization using splitting of middle term method. In this method we are going to split the middle term into two factors. After that we have to take the common factors in brackets and then we will get the factors.

Complete step by step answer:
So, the equation here is \[2{x^2}\, + \,5x\, - \,3\, \leqslant \,0\]. We can also write it as \[2{x^2}\, + \,5x\, - \,3\, = \,0\]. Now, solve it by using the Quadratic factorisation using Splitting of Middle term method. For a polynomial of the form \[a{x^2}\, + \,bx\, + \,c\] , you can rewrite the middle term as the sum of two terms who is having a product as \[a.c\, = \,2. - 3\, = \, - 6\] and who is having a sum as \[b\, = \,5\]
Now, we can factor \[5\,out\,of\,5x\]
\[ \Rightarrow 2{x^2}\, + \,5(x)\, - \,3\, = \,0\]
Now, you have to rewrite \[5\,as\, - 1\,plus\,6\]
\[ \Rightarrow 2{x^2}\, + \,( - 1\, + \,6)(x)\, - \,3\, = \,0\]
By applying the distributive property we get,
\[ \Rightarrow 2{x^2}\, - 1x\, + \,6x\, - \,3\, = \,0\]
Now, we have to factor out the greatest common factor from each group. Here grouping is done by taking the first two terms as the first group and the last two terms as the second group.
\[(2{x^2}\, - 1x)\, + \,(6x\, - \,3)\, = \,0\]
\[ \Rightarrow x(2x\, - 1)\, + \,3(2x\, - \,1)\, = \,0\]
Now, we can factor the polynomial by factoring out the greatest common factor, \[2x\, - \,1\]
\[ \Rightarrow (2x\, - 1)\,(x\, + \,3)\, = \,0\]
Now apply the null method,
\[
(2x\, - 1)\, = \,0 \\
\Rightarrow \,(x\, + \,3)\, = \,0 \\
\]
Now, we can set the factors equal to 0 and solve with the help of null method,
\[\therefore x\, = \,dfrac{1}{2}\,\,and\,\,x\, = \, - 3\]
Since the question was asked whether it was less than equal to \[3\] or not. So, here one of the values is \[ - 3\].

So, in the final solution the value of \[x\,=\dfrac{1}{2}\].

Additional information:
Factorisation is a method in which a mathematical expression or a polynomial is divided into different terms so that it becomes easier to solve the expression. In the factorisation method, we reduce any algebraic or quadratic equation into its simpler form.

Note:Middle term splitting method is easy to find factors, but there is another method that is taking the highest common factor directly and obtaining the factors. This method is working for some questions only, mostly for 3 variable questions. This method makes the question solving very quick and is easy to use.