Oxidation state of Mn in MnO₄⁻

I have tried to understand what the oxidation state for Mn in $$\ce{MnO4^-}$$ is. So far i have understood that O have the oxidation state of -2 so because there is 4 O the number turns to -8. What happens to the minus at O? and what is the oxidation state for Mn?

The sum of the oxidation states of all of the atoms needs to equal the total charge on the ion.

For example, for $$\ce{NH4+}$$, if each hydrogen atom has an oxidation state of +1, and the overall charge is +1, then we can solve for the oxidation state of nitrogen:

\begin{aligned} 4(+1) + OS_\ce{N} &= +1\\ OS_\ce{N} &= +1 -4(+1) \\ OS_\ce{N} &= -3 \end{aligned}

Similarly, we can do $$\ce{SO4^2-}$$. If each oxygen atom has an oxidation state of -2, and the overall charge is -2, we can solve for the oxidation state of sulfur:

\begin{aligned} 4(-2) + OS_\ce{S} &= -2\\ OS_\ce{S} &= -2 -4(-2) \\ OS_\ce{S} &= +6 \end{aligned}

Seeing as Oxygen has an oxidation state of -2 and you have 4 Oxygen atoms, the total charge contributed by Oxygen = -8. Because the total charge of this compound = -1, then in this case Mn = +7 +7 - 8 = -1

As you said Oxygen has the oxidation state of -2 so if we add up all the oxygens they have a -8 oxidation state.

As the oxidation state of the ion is equal to its charge, the Mn has to "balance" the -8 to get it up to -1. Therefore the oxidation state of Mn in Mno4- is +7.