how to balance this redox reaction?

MnO₄

+ C₂O₄

² + H⁺ → Mn⁺² + CO₂ + H₂O

The equation is :
MnO₄^- + C₂H₂O₄ $\to$ Mn²⁺ + CO₂ + H₂O

​The oxidation number of carbon increases from +3 to +4, thus oxidation number increases by 1 per atom or 2 per molecule. The oxidation number of Mn decreases from +7 to +2, thus oxidation number decreases by 5 per atom.

To balance the increase and decrease in oxidation number multiply MnO₄^- by 2 and C₂H₂O₄ by 5.
2MnO₄^- + 5C₂H₂O₄ $\to$ 2Mn²⁺ + CO₂ + H₂O

Balance all atoms except H and O
2MnO₄^- + 5C₂H₂O₄ $\to$ 2Mn²⁺ + 10CO₂ + H₂O

This reaction takes place in acidic medium . The O and H atoms are balanced as:

Balance O atoms by adding equal number of H₂O
2MnO₄^- + 5C₂H₂O₄ $\to$ 2Mn²⁺ + 10CO₂ + 8H₂O

To balance H atoms added H⁺ ions on the side of deficient in H atoms:
2MnO₄^- + 5C₂H₂O₄ +  6H⁺$\to$ 2Mn²⁺ + 10CO₂ + 8H₂O