Answer
The inequality $x^2+3x\geq10$ is valid for values less than -5 and values more than 2 (including them) i.e. $(-\infty,-5]\cap [2,\infty)$.
Work Step by Step
First, simplify:
$x^2+3x\geq10$
$x^2+3x-10\geq0$
$(x+5)(x-2)\geq0$
Now, we find critical points by equating the numerator and denominator to zero:
$(x+5)(x-2)=0$
There are two critical points:
$x_1+5=0\rightarrow x_1=-5$
$x_2-2=0\rightarrow x_2=2$
Next, we are going to take three values: one less than -5; one between -5 and 2; and one more than 2 to test in the original equation and check if the inequality is true or not:
First test with a value less than -5:
$(-6)^2+3(-6)\geq10$
$36-18\geq10$
$18\geq10 \rightarrow \text{ TRUE}$
Second test with a value between -5 and 2:
$0^2+3(0)\geq10$
$0+0\geq10$
$0\geq10 \rightarrow \text{ FALSE}$
Third test with a value more than 2:
$3^2+3(3)\geq10$
$9+9\geq10$
$18\leq10 \rightarrow \text{ TRUE}$
These tests show that the inequality $x^2+3x\geq10$ is valid for values less than -5 and values more than 2 (including them) i.e. $(-\infty,-5]\cap [2,\infty)$.
