Divide.
$$(8x^2+18x+14)÷(2x+3)$$
Your answer should give the quotient and the remainder.

Step 1 ：Given the polynomial division problem $(8x^2+18x+14)÷(2x+3)$.

Step 2 ：We start by dividing the first term of the dividend $8x^2$ by the first term of the divisor $2x$ to find the first term of the quotient, which is $4x$.

Step 3 ：Next, we multiply the divisor $2x+3$ by this term $4x$ and subtract the result from the dividend $8x^2+18x+14$.

Step 4 ：The result of this subtraction is $6x+14$.

Step 5 ：We then repeat the process, dividing the first term of the new dividend $6x$ by the first term of the divisor $2x$, which gives us the next term of the quotient, $3$.

Step 6 ：Multiplying the divisor $2x+3$ by this term $3$ and subtracting the result from the new dividend $6x+14$ gives us the remainder, $5$.

Step 7 ：Since the degree of the remainder is less than the degree of the divisor, we stop here.

Step 8 ：The final quotient is $4x+3$ and the remainder is $5$.

Step 9 ：So, the solution to the polynomial division problem $(8x^2+18x+14)÷(2x+3)$ is the quotient $\boxed{{\displaystyle 4x+3}}$ and the remainder $\boxed{{\displaystyle 5}}$.