Question

find 2 consecutive integers such that three times the square of the first is equal to seven more than five times the second

Answer 1

Answer:

Step-by-step explanation:

We are looking for two numbers, $N$ and $N + 1$. From the problem statement, we can setup the following equation:

$3N^{2} = 5(N + 1) + 7$

$3N^{2} = 5N + 5 + 7$

$3N^{2} - 5N - 12 = 0$

$(3N + 4)(N - 3)$

$N = \frac{-4}{3}, 3$

Because 3 is the only integer solution, the answer is 3 and 4.