Solve 2.00 & 4.60µF Capacitors Voltage: Help Needed

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To find the voltage across the 2.00 µF and 4.60 µF capacitors when connected with opposite signs, the charge (Q) on each capacitor must be calculated using the formula C=Q/V. The total charge will redistribute when the capacitors are connected, affecting the final voltage. The user is struggling to apply the formula correctly and has noted a similar problem with different capacitor values. Assistance is requested to clarify the calculation process. Understanding the charge conservation principle is key to solving this problem effectively.
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Here is the question

A(n) 2.00 µF capacitor is charged to 520 V and a 4.60 µF capacitor is charged to 570 V

What is the voltage for each capacitor if plates of opposite signs are connected?

Any help would be helpful. I have bene trying to get this and haven't been able to all night. I only have one submission left to get the right answer soany suggestions would be greatly appreciated.
 
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The equation is C=Q/V. Just by pluggin in values does not work for me and i can't figure out what I am supposed to do.
 
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