-3•-4=12, and -3+-4=-7.
this will turn into
(x-3)(x-4)
set them up to equal zero
x-3=0, x-4=0
solve
x=3, x=4
(a)
Step-by-step explanation:

Answer:
4480n^3
Step-by-step explanation:
just multiply 80n x 56n and get 4480n but since 80n is to the 2nd and 56n is to the first add them to get to the power of 3 and put that on 4480n
2x - 3y = 8
Subtract '2x' to both sides:
-3y = -2x + 8
Divide -3 to both sides:
y = 2/3x + 8/3
or
y = 2/3x + 2 2/3
a) For this question we set h=59 and solve for t, in order to do so we use the general formula for second-degree equations:
![\begin{gathered} t=\frac{-124\pm\sqrt[]{124^2-4(-16)(-40)}}{2(-16)} \\ t=\frac{-124\pm113.21}{-32} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20t%3D%5Cfrac%7B-124%5Cpm%5Csqrt%5B%5D%7B124%5E2-4%28-16%29%28-40%29%7D%7D%7B2%28-16%29%7D%20%5C%5C%20t%3D%5Cfrac%7B-124%5Cpm113.21%7D%7B-32%7D%20%5Cend%7Bgathered%7D)
The height of the object will be 59 feet at t=7.41 seconds and t=0.34 seconds.
b) When the object reaches the ground, h=0 therefore:

Solving for t we get:
![\begin{gathered} t=\frac{-124\pm\sqrt[]{124^2-4(-16)(19)}}{2(-16)} \\ t=\frac{-124\pm\sqrt[]{16592}}{-32}=\frac{-124\pm128.81}{-32} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20t%3D%5Cfrac%7B-124%5Cpm%5Csqrt%5B%5D%7B124%5E2-4%28-16%29%2819%29%7D%7D%7B2%28-16%29%7D%20%5C%5C%20t%3D%5Cfrac%7B-124%5Cpm%5Csqrt%5B%5D%7B16592%7D%7D%7B-32%7D%3D%5Cfrac%7B-124%5Cpm128.81%7D%7B-32%7D%20%5Cend%7Bgathered%7D)
Therefore, since t cannot be negative the solution is t=7.9 seconds.