How do you solve \[{{x}^{2}}+4x-21=0\]?

Hint: When we have a polynomial of the form \[a{{x}^{2}}+bx+c=0\], we can solve this quadratic with splitting up the b term into two terms based on the signs of a and c terms. If we have the opposite sign for both a and c terms, then we split the b term into two parts. These parts are formed such that the difference of them is b term itself and their product is the same as that of the product of a and c terms.

Complete step by step answer:
In the given quadratic polynomial \[{{x}^{2}}+4x-21\], the coefficient of \[{{x}^{2}}\] and constant terms are of the different sign and their product is -21. That is we have a and c coefficients with opposite signs in the polynomial of form \[a{{x}^{2}}+bx+c\].
Hence, we would split 4, which is the coefficient of \[x\], in two parts, whose sum is 4 and product is -21. These are 7 and -3.
So, we write it as
\[\Rightarrow {{x}^{2}}+4x-21=0\]
We now split \[4x\] into \[7x\] and \[-3x\] .
\[\Rightarrow {{x}^{2}}+7x-3x-21=0\]
We take the terms common in the first 2 terms and last 2 terms.
\[\Rightarrow x(x+7)-3(x+7)=0\]
Here, we have \[(x+7)\] in common then
\[\Rightarrow (x-3)(x+7)=0\]
Now we equate the equation, then we get
\[\begin{align}
  & \Rightarrow x-3=0,x+7=0 \\
 & \Rightarrow x=3,-7 \\
\end{align}\]
\[\therefore x=3,-7\] is the solution of \[{{x}^{2}}+4x-21=0\].

Note:
If sign of coefficient of \[{{x}^{2}}\] and constant terms are same, then we factor the polynomial by splitting up the b term into two terms such that sum of those parts is b term and product is same as that of product of a and c terms. Factoring by grouping will not always work. In such cases we better go with the quadratic formula.