Answer:
{ 2,-2}
Step-by-step explanation:
In order to solve this set we just have to first clear the "q":
[tex]7q^{2} -28=0\\7q^{2} -28+28=+28\\7q^{2} =28\\\frac{7q^{2}}{7} =\frac{28}{7} \\q^{2}=4\\\sqrt{q^{2}} =\sqrt{4} \\q= 2, -2[/tex]
So as we know the solution for a square root is always a negative and positive number, so the solutions ser for 7q2-28=0 is 2 and -2
