Lena and Doc climb into a spaceship. Their trajectory can be modeled by the equation $h=-5/7(d+3)(d-45)$ where $h$ is height above the surface, measured in km, and $d$ is horizontal distance from the launching point, also measured in km. An orbiting spacecraft will fly parallel to the surface of the planet, at a constant height of $65$ km the above the surface. At what distance, to the nearest tenth of a kilometer, will Lena and Doc intercept this spacecraft, if at all? Explain your results. (The speed is not a factor in this problem)
I first made $h=-5/7(d+3)(d-45)$ into standard form, so it became $$h=-5/7d^2+30d+96.42857143.$$ I then let $h=65$, so the previous equation became $$65=-5/7d^2+30d+96.42857143\implies -5/7d^2+30d+31.42857143=0$$
I then used the quadratic formula, which yielded $$d=\frac{-(-30)±\sqrt{30^2-4(-5/7)(31.42857143)}}{2(-5/7)}$$
After doing that I got $d≈43.022$ or $d≈-1.0227$, but since distance can't be negative, the spacecraft and Lena meet at $d=43$.
But I'm not sure if my answer's correct, because I'm wondering if instead of converting the equation into standard form I should have just stuck to the original equation, $h=-5/7(d+3)(d-45)$, and plug $h=65$ into that.
Please let me know which way is correct to find the answer. Thanks.