Solving equations $\left( {2x + 3} \right)\left( {2x + 5} \right)\left( {x - 1} \right)\left( {x - 2} \right) = 30$ the roots available are
A. 0, $\dfrac{1}{2}$, $\dfrac{{11}}{4}$, $\dfrac{9}{4}$
B. 0, $ - \dfrac{1}{2}$, $\dfrac{{ - 1 \pm \sqrt {105} }}{4}$
C. 0, $ - \dfrac{1}{2}$, $ - \dfrac{{11}}{4}$, $ - \dfrac{9}{4}$
D. None

Hint: Here we have been given the polynomial. We will first expand the given polynomial using the distributive property of the multiplication. Then we will factorize the equation and after factoring it, we will get the required roots of the given polynomial.

Formula used:
If $a{x^2} + bx + c = 0$, then the roots of the quadratic equation will be equal to
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.

Complete step by step solution:
 Here we have been given the polynomial i.e. $\left( {2x + 3} \right)\left( {2x + 5} \right)\left( {x - 1} \right)\left( {x - 2} \right) = 30$
We will first expand the given polynomial using the distributive property of multiplication.
$ \Rightarrow 4{x^4} + 4{x^3} - 25{x^2} - 13x + 30 = 30$
Now, we will subtract 30 from both sides.
$ \Rightarrow 4{x^4} + 4{x^3} - 25{x^2} - 13x + 30 - 30 = 30 - 30$
On further simplification, we get
$ \Rightarrow 4{x^4} + 4{x^3} - 25{x^2} - 13x = 0$
Now, we will take the common factor $x$ from all the terms.
$ \Rightarrow x\left( {4{x^3} + 4{x^2} - 25x - 13} \right) = 0$
On further factoring the polynomial equation, we get
$ \Rightarrow x\left( {2x + 1} \right)\left( {2{x^2} + x - 13} \right) = 0$
This is possible when
$x = 0$ ………….. $\left( 1 \right)$
$2x + 1 = 0$ ………… $\left( 2 \right)$
$2{x^2} + x - 13$ ……………. $\left( 3 \right)$
Now, we will further simplify the equation 2.
$ \Rightarrow 2x + 1 = 0$
Now, we will subtract 1 from both sides.
$
   \Rightarrow 2x + 1 - 1 = 0 - 1 \\
   \Rightarrow 2x = - 1 \\
$
Now, we will divide 2 on both sides.
$
   \Rightarrow \dfrac{{2x}}{2} = \dfrac{{ - 1}}{2} \\
   \Rightarrow x = \dfrac{{ - 1}}{2} \\
$
Now, we will simplify the equation 3.
$2{x^2} + x - 13 = 0$
We can see that this is a quadratic equation.
We will directly apply the formula to find the roots.
We know if $a{x^2} + bx + c = 0$, then the roots of the quadratic equation will be equal to
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Using this formula of roots, we get
$ \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {{1^2} - 4 \times 2 \times - 13} }}{{2 \times 2}}$
On further simplification, we get
$ \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {1 \pm 105} }}{4}$
Hence, the required roots of the given polynomial are:-
0, $ - \dfrac{1}{2}$ and $\dfrac{{ - 1 \pm \sqrt {105} }}{4}$

Hence, the correct option is option B.

Note: here we have used the formula to find the roots of the quadratic equation obtained after factoring the polynomial. Here roots of the quadratic equation are defined as the values which when put in the place of the variables will satisfy the equation. The number of roots of the polynomial is always equal to the degree of that polynomial.