How do you factor $ 3{{x}^{2}}+14x-5=0 $ ?

Hint: We will first take the product of the constant term and the leading coefficient of the given equation. Then we will look at a list of some of the pairs of factors of this product. We will find the pair of factors that give a difference equal to the coefficient of the middle term. We will then split the given equation into factors by modifying the given equation.

Complete step by step answer:
The given equation is $ 3{{x}^{2}}+14x-5=0 $ . The leading coefficient of the equation is 3 and the constant term is $ -5 $. The product of the leading coefficient and the constant term of the equation is $ 3\times -5=-15 $. Let us look at some pair factors of the number 15. We know that $ 5\times 3=15 $ and $ 15\times 1=15 $ . The middle term of the given equation has a coefficient of 14. We can see that $ 15-1=14 $ and $ 15\times -1=-15 $ .
Now, we will split the middle term of the given equation in the following manner,
$\Rightarrow$ $ 3{{x}^{2}}+15x-x-5=0 $
We can take out $ 3x $ as a common factor from the first two terms and $ -1 $ from the third and the fourth term. So, the above equation will look like the following,
$\Rightarrow$ $ 3x\left( x+5 \right)-1\left( x+5 \right)=0 $
Now, we can take out the term $ \left( x+5 \right) $ as a common factor. So, we get the following equation,
$\Rightarrow$ $ \left( x+5 \right)\left( 3x-1 \right)=0 $
We have obtained an equation as a product of two factors. Therefore, this equation is the complete factorization of the given quadratic equation.

Note:
We can solve the quadratic equation by equating either of the two factors to zero. So, the solution of the quadratic equation is $ x=-5 $ and $ x=\dfrac{1}{3} $ . It is important to understand the concept of factorization. It is very useful in simplifying or solving quadratic equations. There are other methods like completing the square or quadratic formula method to solve quadratic equations.