Find the quotient and the remainder when $2x^{2}-7x-15$ is divided by $2x+1$.

(N/A) To find the quotient and remainder,we perform polynomial long division:
$1$. Divide the first term of the dividend $(2x^2)$ by the first term of the divisor $(2x)$: $2x^2 / 2x = x$. This is the first term of the quotient.
$2$. Multiply the divisor $(2x+1)$ by $x$: $x(2x+1) = 2x^2 + x$.
$3$. Subtract this from the dividend: $(2x^2 - 7x - 15) - (2x^2 + x) = -8x - 15$.
$4$. Divide the first term of the new expression $(-8x)$ by the first term of the divisor $(2x)$: $-8x / 2x = -4$. This is the second term of the quotient.
$5$. Multiply the divisor $(2x+1)$ by $-4$: $-4(2x+1) = -8x - 4$.
$6$. Subtract this from the current expression: $(-8x - 15) - (-8x - 4) = -8x - 15 + 8x + 4 = -11$.
Therefore,the Quotient is $x-4$ and the Remainder is $-11$.