Lottery +EV point?

This state lottery changed the format for their Cash 5 game:
https://www.palottery.state.pa.us/Draw-Games/Cash-5/Prizes-Chances.aspx

Used to be $1 for a ticket thus if the jackpot was over $962,598, then it's +EV, assuming you're the only winner. (For this exercise, ignore taxes)

Now it's $2 for a ticket and they added a side game gimmick called 'Quick Cash' for the extra dollar.
Overall odds of winning that is 1:4.71

1/4.71 = .21
1-.21= .79

.79 x 962,598 = 760,452
962,598 + 760,452 = 1,723,050

so if the jackpot is greater than $1,723,050 then it's +EV? (again assuming you're the only winner and ignore taxes)

1) If not, then what's the answer?
2) And where did my math go wrong?

I get a breakeven jackpot of $1,190,941.95*.

That was a poorly written rules page you linked to, in my opinion. Here is some more needed to calculate the odds:

1. 43 total balls. Player picks 5.
2. As you mention, the ticket cost is $2.
3. Game pays jackpot for matching 5, $200 for matching 4, $10 for matching 3, and $2 for matching 2.
4. I assume the player gets something like a scratch-off with the pick-5 game. This scratch-off has a 1/80 chance of a win of $6 and 1/5 chance of a win of $2.

* Corrected

Quote: Wizard

I get a breakeven jackpot of $558,241.95.

How do you get this?
Here's what I get:
Each $2 ticket has two components - a Cash 5 component, and a Quick Cash component.
The Quick Cash component has an EV of 1/5 x 2 + 1/80 x 6 = 19/40
The Cash 5 component's EV is the sum of:
962,598
2 numbers: 2 x 84,360 / 962,598 = 168,720 / 962,598
3 numbers: 10 x 7030 / 962,598 = 70,300 / 962,598
4 numbers: 200 x 190 / 962,598 = 38,000 / 962,598
5 numbers: 1 / 962,598 x (the size of the jackpot) / (the expected number of jackpot winners)
If there are N tickets sold, the expected number of jackpot winners is N / 962,598, so the 5 numbers EV value = 1 / 962,598 x J / (N / 962,598) = J / N.
The total EV = 19/40 + 277,020 / 962,598 + jackpot size / number of tickets
According to the rules, the jackpot size = the previous jackpot size + N x 0.390632 (44.39% of each ticket's price goes into the prize pool, of which 44% goes into the jackpot)
The total EV = 1.1534157 + the previous jackpot amount / the number of tickets bought

In my original math I put a win of $100 for matching 4. It is supposed to be only $10. This causes me to change my answer is $1,190,941.95.

Ahh.. didn't take the dollar amount into consideration when calculating the quick cash. (Thx)
19/40 = 47.5% return

.475 x 962,598 = 457,234

Taking only the Cash5 top prize into consideration (because that's the only prize people care about) then:
457,234 + 962,598 = $1,419,832

That's the point to be playing at?

Here's another game (Match 6):
https://www.palottery.state.pa.us/Draw-Games/Match-6.aspx
https://www.palottery.state.pa.us/Draw-Games/Match-6.aspx#howtoplay

$2 per ticket, you get 3 plays. (so you get 18 numbers per $2 ticket)
odds of winning the top prize is 1 in 4,661,272 in a 3 play panel.

but they also have a secondary game gimmick.
if you match 4 to 10 numbers out of the 18, you also win some $.
the more #s you match, the more $ you get.

What's the +EV point of this game?
Thx