To calculate: The partial decomposition of the function \frac{-12x-

Given Information:
The provided expression is ${\displaystyle \frac{-12x-29}{2x^2+11x+15}}$.
Formula used:
Decomposition of ${\displaystyle \frac{f\left(x\right)}{g\left(x\right)}}$ into Partial Fractions:
Consider a rational expression ${\displaystyle \frac{f\left(x\right)}{g\left(x\right)}}$, where f(x) and g(x) are polynomial with real coefficients, ${\displaystyle g\left(x\right)\ne 0}$, and the degree of f(x) is less than degree of g(x).
Step 1: Factor the denominator g(x) completely into linear factors of the form ${\displaystyle {(ax+b)}^m}$ and quadratic factors of the form ${\displaystyle {(a{x}^2+bx+c)}^n}$ that are not further factorable over the integers.
Step 2: Set up the form of decomposition. That is, write the original rational expression ${\displaystyle \frac{f\left(x\right)}{g\left(x\right)}}$ as a sum of simpler fractions using these guidelines. Note that ${\displaystyle A_1,A_2,\dots .A_m,B_1,B_2,\dots .,B_m}$ and ${\displaystyle C_1,C_2,\dots .C_m}$ are constants.
Linear Factors of g (x):
For each linear factor of g(x), the partial fraction decomposition must include the sum:
${\displaystyle \frac{A_1}{{(ax+b)}^1}+\frac{A_{\{}}{{(ax+b)}^2}+\dots +\frac{A_m}{{(ax+b)}^m}}$
Quadratic Factors of g(x):
For each quadratic factor of g(x), the partial fraction decomposition must include the sum:
${\displaystyle \frac{B_{1}x+C_1}{{(a{x}^2+bx+c)}^1}+\frac{B_{2}x+C_2}{{(a{x}^2+bx+c)}^2}+\dots +\frac{B_{n}x+C_n}{{(a{x}^2+bx+c)}^n}}$
Step 3: With the form of the partial fraction decomposition set up, multiply both sides of the equation by the Least Common Divisor to clear fractions.
Step 4: Use the equation from step 3, set up a system of linear equations by equating the constant terms and equating the coefficients of like powers of x.
Step 5: Solve the system of equations from step4 and substitute the solutions to the system into the partial fraction decomposition. 
Calculation:
Consider the provided expression,
${\displaystyle \frac{-12x-29}{2x^2+11x+15}}$
Here, ${\displaystyle f\left(x\right)=-12x-29}$ and ${\displaystyle g\left(x\right)=2x^2+11x+15}$
Factorize g(x):
${\displaystyle g\left(x\right)=2x^2+11x+15}$
${\displaystyle =2x^2+6x+5x+15}$
${\displaystyle =2x(x+3)+5(x+3)}$
${\displaystyle =(2x+5)(x+3)}$
So, the expression can be written as: