How do you solve the inequality \[{{x}^{2}}+3x>10\]?

Hint: This type of problem is based on the concept of inequality. First, we have to consider the whole function and then add -10 on both the sides of the inequality. And convert the right-hand side of the inequality to zero. Then, we need to find the factors of the given inequality by making necessary calculations. And then solve x by making some adjustments to the given inequality.

Complete step by step answer:
According to the question, we are asked to solve the given inequality \[{{x}^{2}}+3x>10\].
We have been given the inequality is \[{{x}^{2}}+3x>10\] . -----(1)
We first have to add -10 on both the sides of the equation (1).
\[\Rightarrow {{x}^{2}}+3x-10>10-10\]
On further simplifications, we get,
\[{{x}^{2}}+3x-10>0\] --------(2)
We know that
3x=5x-2x
And -10=5(-2)
Therefore, substituting these observations in equation (2), we get
\[{{x}^{2}}+5x-2x+5\left( -2 \right)>0\]
By taking x common from the first two terms of the inequality and -2 common from the last two terms of the inequality, we get
\[\Rightarrow x\left( x+5 \right)-2\left( x+5 \right)>0\]
Now take \[\left( x+5 \right)\] common from the obtained inequality.
We get,
\[\Rightarrow \left( x+5 \right)\left( x-2 \right)>0\]
Since \[\left( x+5 \right)\left( x-2 \right)>0\], \[\left( x+5 \right)\] and \[\left( x-2 \right)\] are greater than 0.
\[\left( x+5 \right)>0\] and \[\left( x-2 \right)>0\].
\[\Rightarrow x+5-5>0-5\] and \[x-2+2>0+2\]
\[\therefore x>-5\] and \[x>2\]
Hence, the values of x in the inequality \[{{x}^{2}}+3x>10\] are \[x>-5\] and \[x>2\].

Note:
Whenever you get this type of problem, we should always try to make the necessary changes in the given inequality to get the final solution of the inequality which will be the required answer. We should avoid calculation mistakes based on sign conventions. We should always make some necessary calculations to obtain zero in the right-hand side of the inequality. And then solve the inequality by factoring the quadratic equation with the variable x.