\(T_1 =\frac{28}{14-W}\)

\(T_2=\frac{28}{14+W}\)

\(T_1-T_2=\frac{35}{60}\)

\(\frac{28}{14-W} - \frac{28}{14+W} = \frac{35}{60}\) In this stage try to plug in the numbers (or POE) and get the answer quickly

\(\frac{1}{14-W} - \frac{1}{14+W} = \frac{35}{60*28}=\frac{1}{48}\)

\(\frac{14+W-(14-W)}{(14^2-W^2)}=\frac{1}{48}\)

\(48*2W=196-W^2\)

\(W^2+96W-196=0\)

\(W^2+98W-2W-196=0\)

\((W+98)(W-2)=0\)

\(W=2,-98\)

Ans E

Solution:

Letting r be the speed of the stream in kmph and t be the time, in hours, for the boat to travel upstream, we can create the equations:

(14 - r) * t = 28

and

(14 + r) * (t - 35/60) = 28

From the first equation, we see that t = 28 / (14 - r). Substituting this into the second equation, we have:

(14 + r) * (28 / (14 - r) - 7/12) = 28

Multiplying the equation by 12(14 - r), we have:

(14 + r) *[(28 * 12 - 7(14 - r)] = 28 * 12(14 - r)

Dividing the equation by 7, we have:

(14 + r) * (4 * 12 - (14 - r)) = 4 * 12(14 - r)

(14 + r)(34 + r) = 48(14 - r)

476 + 48r + r^2 = 672 - 48r

r^2 + 96r - 196 = 0

(r - 2)(r + 98) = 0

r = 2 or r = -98

Since r can’t be negative, r = 2.

Answer: E
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I was stuck at x^2+96x-196=0, thanks for the explanations

I decided to take a little different approach. Since we know that in still water boat speed is 14k/h then time needed to swim for 28km is 2 hr. Now we can make two equations
28= (14 + x )* (2hr -35 min ) Speed downstream
28= (14 - x )* (2hr +35 min ) Speed upstream

The most work now is in these brackets (2hr -35 min ), (2hr +35 min ). In order to simplify we can just use 30 min instead of 35. so we have
28= (14 + x )* (1.5 hr )
28= (14 - x )* (2.5 hr )

(14 + x )* (1.5 hr )=(14 - x )* (2.5 hr )
21+1.5x = 35 -2.5x
4x=14
x= 14/4
x=3.5 km/h
Since we reduced 35 min down to 30 min we know that the speed of stream is less that 3.5 km/h
And now we can see that the only option is 2km/h

P.S. I understand that this is not the best way to solve such question, but this is pretty fast and simple way to choose proper answer in question such this.