binary equation

Author: qiwsir

newcodes/answers/q53.py

@@ -0,0 +1,31 @@
+#!/usr/bin/env python
+# coding=utf-8
+
+import math
+import cmath
+
+def solve_be(a, b, c):
+    delta = b**2 - 4*a*c
+    if delta == 0:
+        #x = fractions.Fraction(-b, 2*a)
+        x = -b / (2*a)
+        return x
+    elif delta > 0:
+        sqrt_delata = math.sqrt(delta)
+    else:
+        sqrt_delata = cmath.sqrt(delta)
+    #x1 = fractions.Fraction((-b + sqrt_delata), 2*a)
+    #x2 = fractions.Fraction((-b - sqrt_delata), 2*a)
+    x1 = (-b + sqrt_delata) / (2*a)
+    x2 = (-b - sqrt_delata) / (2*a)
+    return (x1, x2)
+
+if __name__ == "__main__":
+    print("The binary linear equation is x^2 + 2^x + 3 = 0")
+    r = solve_be(1, 2, 3)
+    if len(r) == 1:
+        print("The equation only has one root. It is:")
+        print(r)
+    else:
+        print("The equation have two root. They are:")
+        print(r)