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aX^2 + bX + c
Ax^2 + bx + c = 0
Ax 2+Bx+c=0
Solve the quadratic equation.If y = ax^2+bx+c then the intercepts arex = [-b +/- sqrt(b^2 - 4ac)]/(2a).The solutions are real if and only if b^2 >= 4ac
Without using complex numbers, which is a bad idea, you can't. But you can do something which is almost as good: instead of dealing with -ax^2+bx+c, you can take -(ax^2-bx-c), and then complete the square for the inside quadratic. This will give you something like -((rx+s)^2+t), which is usually good enough. For example, if you're completing the square in order to solve a quadratic equation, the sign on the outside doesn't matter: if -((rx+s)^2+t)=0 then (rx+s)^2+t=0 as well (because the negative of 0 is 0). A different approach, which is usually more useful, is to factor out the a, regardless of whether it's positive or negative. Then instead of completing the square for ax^2+bx+c, you complete the square for x^2+b/ax+c/a, and multiply the final result by a.