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

Hint: When we factorize a quadratic equation $a{{x}^{2}}+bx+c$ we write $bx$ as $mx+nx$ where $m+n=b$ and $mn=ac$, then we can easily factorize the equation. We can split 14x to 15x and –x to solve this question.

Complete step-by-step answer:
The given equation is $5{{x}^{2}}+14x-3$ which is a quadratic equation. if we compare the equation to standard quadratic equation $a{{x}^{2}}+bx+c$ then a = 5, b = 14, and c = -3
To factor a quadratic equation we can find two numbers m and n such that the sum of m and n is equal to b and the product of m and n is $ac$. Then we can split $bx$to $mx+nx$ then we can factor the equation easily.
In our case
ac = -15
and b = 14
So pair of 2 numbers whose product is - 15 are (1,- 15) , (3,- 5) , (-3, 5) and (-1, 15)
But there is only one pair whose sum is 14
We can $6x$ spilt to $15x-x$
So $5{{x}^{2}}+14x-3$ = $5{{x}^{2}}+15x-x-3$
Taking 5x common in the first half of the equation and taking - 1 common in the second half of the equation.
$\Rightarrow 5{{x}^{2}}+14x-3=5x\left( x+3 \right)-1\left( x+3 \right)$
Taking x + 3 common
$\Rightarrow 5{{x}^{2}}+14x-3=\left( 5x-1 \right)\left( x+3 \right)$
In this way we can factor a quadratic equation; another method is directly find the roots of the equation.

Note: While factoring a quadratic equation we can’t always split $bx$ such that their product is equal to ac because sometimes the roots can be irrational numbers. So in that case we can try the complete square method to factorize the equation.