Solve for Time: Expanding (40a+30)(4a+5)-746 in Planetary Hours

To solve this problem, we'll determine the number of hours in a day on this planet:

• Step 1: Calculate the total number of minutes in a day given as $(40a + 30)(4a + 5) - 746$.
• Step 2: The total number of minutes calculated per the day's duration is $(20a - 4) \times (8a + 3)$ since there are $20a - 4$ hours each lasting $8a + 3$ minutes.
• Step 3: Equate the two expressions.

The given expression for the total minutes is:

$(40a + 30)(4a + 5) - 746$

Calculate the expression without subtraction:

$(40a + 30)(4a + 5) = 40a \times 4a + 40a \times 5 + 30 \times 4a + 30 \times 5$ $= 160a^2 + 200a + 120a + 150$ $= 160a^2 + 320a + 150$

Subtract 746:

$160a^2 + 320a + 150 - 746 = 160a^2 + 320a - 596$

The total minutes also correspond to:

$(20a - 4)(8a + 3)$ $= 20a \times 8a + 20a \times 3 - 4 \times 8a - 4 \times 3$ $= 160a^2 + 60a - 32a - 12$ $= 160a^2 + 28a - 12$

Now equate the expressions:

$160a^2 + 320a - 596 = 160a^2 + 28a - 12$

Subtract $160a^2$ from both sides:

$320a - 596 = 28a - 12$

Subtract $28a$ from both sides and add 596 to both sides:

$320a - 28a = -12 + 596$ $292a = 584$

Solve for $a$:

$a = \frac{584}{292} = 2$

Knowing $a = 2$, calculate hours per day:

$20a - 4 = 20(2) - 4 = 40 - 4 = 36$

Therefore, the number of hours in a day is $2$ hours.

The correct answer is: 2 hours.