One more math problem!!! please help me out. thanks! i know the answers but i need the work!!!?

Question

Suppose your car contains just one gallon of gas. Driving at 20 mi/h you can go 26 mi. Likewise, you can go 34 mi driving at 40 mi/h and 32 mi driving at 50 mi/h.

a. find a quadratic function that models this data
b. how far could you go if you drove at 65 mi/h
c. the nearest gas station is 16mi away. if the speed limit is 55mi/h, at what maximum speed could you drive and still reach it?

Answer 1

distance = ax^2 +bx+c where x is the velocity
putting in the distances and velocities
26=400a+20b+c
34 =1600a+40b+c
32=2500a +50b+c
8= 1200a+20b
-2=900a +10b and -4=1800a+20b
-12=600a and a= -1/50
-2= -18+10b so b= 1.6
and 26+8-32=c=2

d= -1/50 x^2+1.6 x+2
d=(-1/50)*65^2 +1.6*65 +2 =21.5 miles
16 = -1/50x^2+1.6 x+2
800 = -x^2 +80x +100 so x^2-80x +700 =0
x =((80+-sqrt( 6400-2800)/2 = 40+-30 miles/h
The sum= 70 ml/h is beyond the speed limit so you must drive at rhe speed limit

Answer 2

y = ax^2+bx+c
26 = 400a +20b +c <-- Eq1
34 = 1600a +40b +c <-- Eq 2
32 = 2500a +25b +c <-- Eq 3
-8 = -1200a -20b <--EQ4 = Eq 1-Eq 2
2 = -900a +15b <-- Eq 5= EQ2-Eq 3
-24 = -3600a -60b <-- Eq6 = 3*Eq 4
8 = -3600a +60b <-- Eq 7 = 4*EQ5
-16 = -7200a
a= 16/7200 = 1/450
2= -900/450 +15b
4= 15b --> b = 4/15
26 = 400/450 +20(4/15)+c
26 = 8/9 + 16/3 + c --> c = 26-56/9 = 19 7/9
So equation is y = x^2/450 +4x/15 + 19 7/9

Something wrong. a should be negative so we get a maximum. Anyhow, this is the way to solve it.