Python – Breaking out of a deeply nested loop with `while`, `try`, `if`, and `except` statements

Here is my code that solves a quadratic equation given user inputs a, b, c. However, I want to check for incorrect user inputs. If a user inputs anything but a float, the user is confronted with print "Not a valid input. Try again.\n" followed by a loop of the function quadratic2(). However, this program is part of a larger program, so I want the user to have the option of typing in "next" to terminate this function and move on. My issue is that I am calling for 3 user inputs, but I want the "next" input to be the first input, a. This works perfectly well until the user types in a random string, causing a ValueError. Once this happens, my code does not want to loop back to check for if the input is next. Please help me! I think something about the way this code is nested is messing me up.

def retest3():
    print "Type in another a, b, and c. Or type \"Next\" to move on."
def quadratic1():
    print ("This program calculates the zeroes of a quadratic equation."
    "\nPlease enter a, b, and c, hitting \"Enter\" after each one: ")
def quadratic2():
    while 1 == 1:
        a = (raw_input())
        if "next" == a.lower():
            ontonextthing()
            return 1 == 0
        else:
            try:
                a = float(a)
            except ValueError:
                print "Not a valid input. Try again.\n"
                quadratic2()
            try:
                b = float(raw_input())
            except ValueError:
                print "Not a valid input. Try again.\n"
                quadratic2()
            try:
                c = float(raw_input())
            except ValueError:
                print "Not a valid input. Try again.\n"
                quadratic2()

            if b**2-4*a*c>0:
                root1=(-b+math.sqrt(b**2-4*a*c))/(2*a)
                root2=(-b-math.sqrt(b**2-4*a*c))/(2*a)
                print ("First Root: {0}\nSecond Root: {1}".format(root1,root2))
            else:
                print ("The discriminant is negative, so your"
                " roots will contain \"i\".")
                disc1=(b**2-4*a*c)
                disc2=-disc1
                sqrtdisc2=(math.sqrt(disc2))/(2*a)
                b2=(-b)/(2*a)
                print ("{0} + {1}i".format(b2, sqrtdisc2))
                print ("{0} - {1}i\n".format(b2, sqrtdisc2))
        retest3()   

quadratic1()
quadratic2()