Sorry if I have to copy past the question but I can't understand where I'm wrong in this specific problem:
Two drivers are testing the same model car at speeds that differ by 20 km/hr. The one driving at the slower rate drives 70 kilometers down a speedway and returns by the same route. The one driving at the faster rate drives 76 kilometers down the speedway and returns by the same route. Both drivers leave at the same time, and the faster car returns $\frac{1}{2}$ hour earlier than the slower car. At what rates were the cars driven?
I get two equations from this problem which is: $(v-20)(t)=70$ and $(v)(t-\frac{1}{2})=76$
after doing substitution I get to the equation: $v^2-8v-3040=0$ which gives faster car 59.28 and slower car 39.28 but the answers in textbook is: 56 and 76. So am I wrong or this is just an error in the book?
As player3236 has noted, the distances should be doubled, as the total distance is the distance to there and back.
Using speed $\times$ time = distance like before, we should have:
$$(v-20)(t) = 140 \Rightarrow t = \frac{140}{v-20} \tag{1}$$ $$(v)(t - 1/2) = 152 \Rightarrow t = \frac{152}{v}+\frac{1}{2} \tag{2}$$
Equating $(1)$ and $(2)$, we find that:
$$\frac{140}{v-20} (v)(v-20) = \frac{152}{v} (v)(v-20) + \frac{1}{2}(v)(v-20)$$ $$\Rightarrow 140v = 152(v-20) + \frac{1}{2}v(v-20)$$ $$\Rightarrow 280v = 304(v-20) + v(v-20)$$ $$\Rightarrow 280v = 304v - 6080 + v^2 - 20v$$ $$\Rightarrow v^2 + 4v - 6080 = 0$$
Use the quadratic formula to find that $v = 76, -80$. Now discard the negative solution, so the faster car has a speed of $76$ km/h, and the slower car is $20$ km/h slower, which is $56$ km/h.