Solve the following quadratic equation by factorization, the roots are $5, - \dfrac{1}{3}$, $3{x^2} - 14x - 5 = 0$
$A)$ True
$B)$ False

Hint: The word quadratic means second-degree values of the given variables.
Then by further simplifying, the quadratic equation $a{x^2} + bx + c = 0$ can be written in the form of $(x - \alpha )(x - \beta ) = 0$.
In this equation, the factors of the quadratic equation are$(x - \alpha )$ and $(x - \beta )$.
So, to determine the roots of the equation, these factors of the equation will be made equal to $0$.
That is,
$(x - \alpha ) = 0$ $(x - \beta ) = 0$
Which gives,
$x = \alpha $ $x = \beta $
So, $\alpha $and $\beta $are the roots of the quadratic equation.
Here we are asked to solve the given quadratic equation, that is we have to find its roots. Since it is an equation of order two it will have two roots. The roots of a quadratic equation can be found by using the formula.

Complete step-by-step solution:
It is given that $3{x^2} - 14x - 5 = 0$ we aim to solve this equation, that is we have to find its roots.
We know that the number of roots of an equation is equal to its degree. Here the degree of the given equation is two thus this equation will have two roots.
Since given that $3{x^2} - 14x - 5 = 0$. Now convert the value $ - 14x = - 15x + x$ then we get $3{x^2} - 15x + x - 5 = 0$
Now taking the common values out, we have $3{x^2} - 15x + x - 5 = 0 \Rightarrow 3x(x - 5) + 1(x - 5) = 0$
Again, using the multiplication operation, take the common multiples then we get $3x(x - 5) + 1(x - 5) = 0 \Rightarrow (x - 5)(3x + 1) = 0$
Thus, we get $x - 5 = 0$ and $3x + 1 = 0$
Hence, we get $x = 5$ and $x = - \dfrac{1}{3}$ which is the required answer and it is true.
Therefore, the option $A)$ True is correct.

Note: The quadratic equation $a{x^2} + bx + c = 0$ can be factored in the form of $(x - \alpha )(x - \beta ) = 0$ where $(x - \alpha )$ and $(x - \beta )$are the factors of the quadratic equation so $\alpha $and $\beta $are the roots of the quadratic equation and the values of $a$ cannot be $0$.
If the value of $a$ is $0$, then the quadratic equation will become a binomial equation.
We can also able to make use of the formula of the quadratic equation. Let \[a{x^2} + bx + c = 0\] be a quadratic equation then the roots of this equation are given by \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].