Well I will put an example but the easiest way
Expanding-
When u ex and
-30x-12y+36x-12y
Combine the x' s first
6x-12y-12y
Combine the y's
6x-24y is your answer
Answer:

Step-by-step explanation:
The motion equations that describe the ball are, respectively:
![x = \left[\left(152\,\frac{ft}{s} \right)\cdot \cos 52^{\circ} \right] \cdot t](https://tex.z-dn.net/?f=x%20%3D%20%5Cleft%5B%5Cleft%28152%5C%2C%5Cfrac%7Bft%7D%7Bs%7D%20%5Cright%29%5Ccdot%20%5Ccos%2052%5E%7B%5Ccirc%7D%20%5Cright%5D%20%5Ccdot%20t)
![y = 4.5\,ft + \left[\left(152\,\frac{ft}{s} \right)\cdot \sin 52^{\circ} \right] \cdot t - \frac{1}{2}\cdot \left(32.174\,\frac{ft}{s^{2}} \right) \cdot t^{2}](https://tex.z-dn.net/?f=y%20%3D%204.5%5C%2Cft%20%2B%20%5Cleft%5B%5Cleft%28152%5C%2C%5Cfrac%7Bft%7D%7Bs%7D%20%5Cright%29%5Ccdot%20%5Csin%2052%5E%7B%5Ccirc%7D%20%5Cright%5D%20%5Ccdot%20t%20-%20%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20%5Cleft%2832.174%5C%2C%5Cfrac%7Bft%7D%7Bs%5E%7B2%7D%7D%20%5Cright%29%20%5Ccdot%20t%5E%7B2%7D)
The time required for the ball to hit the ground is computed from the second equation. That is to say:
![4.5\,ft + \left[\left(152\,\frac{ft}{s} \right)\cdot \sin 52^{\circ} \right] \cdot t - \frac{1}{2}\cdot \left(32.174\,\frac{m}{s^{2}} \right) \cdot t^{2} = 0](https://tex.z-dn.net/?f=4.5%5C%2Cft%20%2B%20%5Cleft%5B%5Cleft%28152%5C%2C%5Cfrac%7Bft%7D%7Bs%7D%20%5Cright%29%5Ccdot%20%5Csin%2052%5E%7B%5Ccirc%7D%20%5Cright%5D%20%5Ccdot%20t%20-%20%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20%5Cleft%2832.174%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%20%5Cright%29%20%5Ccdot%20t%5E%7B2%7D%20%3D%200)
Given that formula is a second-order polynomial, the roots of the equation are described below:
and 
Just the first root offers a realistic solution. Then,
.
b=3q/8 or at least thats what I got
Answer:
Tn=a+(n-1)d
a=a1
Tn=a1+(n+1)d
I would appreciate if my answer is chosen as a brainliest answer