有人可以检查一下我这个初学者 C 程序的算术吗？
我正在用 C 编写一个程序来计算以下公式：
（来源：crewtonramoneshouseofmath.com）
这是行代码（我只是使用+而不是+-）：
x = ((-1 * b) + (sqrt(pow(b, 2) - 4 * a * c)))/(4 * a);
我没有得到正确的根。例如，如果 a = 1、b=-2 和 c=-2，则它应该为 2.73。相反，我得到的是 1.37。
一直盯着代码，我没有看到错误。有人可以帮我指出吗？
I am writing a program in C that calculates this formula:
(source: crewtonramoneshouseofmath.com)
here is the line of code (I am just using + instead of the +-):
x = ((-1 * b) + (sqrt(pow(b, 2) - 4 * a * c)))/(4 * a);
I am not getting the correct root. For example if a = 1, b=-2, and c=-2 it SHOULD be 2.73. Instead I am getting 1.37.
Been staring at the code and I don't see the mistake. Can someone point it out for me?
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评论（5）
x = (...) / (4 * a)这不应该是
2 * a吗？x = (...) / (4 * a)Shouldn't this be
2 * a?有趣的是，1.37（你得到的结果）大约是 2.73（你想要的结果）的一半，你瞧，它就在你的分母中，除以
4a而不是2a。就我个人而言，我会将该表达式写为：
因为它更接近于您尝试复制的方程（
-1 * b更好地表示为-b）并且我发现调用pow来获得一个简单的正方形是不必要的，而b * b在没有函数调用的情况下完成相同的工作。It's interesting that 1.37 (what you're getting) is about half of 2.73 (what you want) and, lo and behold, there it is in your denominator, dividing by
4ainstead of2a.Personally, I would write that expression as:
since it more closely matches the equation you're trying to duplicate (the
-1 * bis better expressed as-b) and I find callingpowto get a simple square to be unnecessary whereb * bdoes the same job without a function call.x1 = ((-1 * bcoeff) + (sqrt(pow(bcoeff, 2) - 4 * acoeff * ccoeff)))/(2 * acoeff);
x1 = ((-1 * bcoeff) + (sqrt(pow(bcoeff, 2) - 4 * acoeff * ccoeff)))/(2 * acoeff);
在这里检查一下自己：
.../(4 * acoeff)Check yourself here:
.../(4 * acoeff)虽然这个问题已经得到解答并指出了您的错误。我只想提一下，将问题分解为更多行会增加其可读性并可能减少所犯的错误。
While this question has already been answered and your error pointed out. I would just like to mention that breaking the problem down into more lines would increase its readability and potentially reduce mistakes made.