How do you solve the inequality ${{x}^{2}}+2x-3\le 0$?

Hint: The given inequation can be solved by taking the quadratic equation and factoring it. We use both grouping methods and vanishing methods to find the factor of the problem. We take common terms out to form the multiplied forms. From there we find the interval which is suitable for the inequation.

Complete step by step answer:
We apply the middle-term factoring or grouping to factorise the polynomial.
Factorising a polynomial by grouping is to find the pairs which on taking their common divisor out, give the same remaining number.
In case of ${{x}^{2}}+2x-3$, we break the middle term $2x$ into two parts of $3x$ and $-x$.
So, ${{x}^{2}}+2x-3={{x}^{2}}+3x-x-3$. We have one condition to check if the grouping is possible or not. If we order the individual elements of the polynomial according to their power of variables, then the multiple of end terms will be equal to the multiple of middle terms.
Here multiplication for both cases gives $-3{{x}^{2}}$. The grouping will be done for ${{x}^{2}}+3x$ and $-x-3$
We try to take the common numbers out.
For ${{x}^{2}}+3x$, we take $x$ and get $x\left( x+3 \right)$.
For $-x-3$, we take $-1$ and get $-\left( x+3 \right)$.
The equation becomes ${{x}^{2}}+2x-3={{x}^{2}}+3x-x-3=x\left( x+3 \right)-\left( x+3 \right)$.
Both the terms have $\left( x+3 \right)$ in common. We take that term again and get
$\begin{align}
  & {{x}^{2}}+2x-3 \\
 & =x\left( x+3 \right)-\left( x+3 \right) \\
 & =\left( x+3 \right)\left( x-1 \right) \\
\end{align}$
So, we get $\left( x+3 \right)\left( x-1 \right)\le 0$.
This gives $x\ge -3$ and $x\le 1$. The interval will be $-3\le x\le 1$.

Note:
We can solve the inequation $\left( x+3 \right)\left( x-1 \right)\le 0$ by taking their signs in consideration. Multiplication of two terms is negative. Then it is essential for two terms to have opposite signs.
The choices are $x\le -3,x\ge 1$ which is not possible and $x\ge -3,x\le 1$ which gives $-3\le x\le 1$.