Asked by Sofa
                x^2 +b/a +(b/2a)^2 = -c/a + (b/2a)^2
            
            
        Answers
                    Answered by
            oobleck
            
    The (b/2a)^2 goes away, since it's on both sides. That leaves (after fixing your typo)
x^2 + bx/a = -c/a
ax^2 + bx = -c
ax^2 + bx + c = 0
However, it appears you want to solve by completing the square
x^2 + b/a x + (b/2a)^2 = (x + b/2a)^2
So that gives
(x + b/2a)^2 = -c/a + b^2/4a2
(x + b/2a)^2 = (b^2-4ac)/(2a)^2
now take square roots
x + b/2a = ±√(b^2-4ac)/2a
x = (-b±√(b^2-4ac)) / 2a
which is the quadratic formula for solving the first equation
    
x^2 + bx/a = -c/a
ax^2 + bx = -c
ax^2 + bx + c = 0
However, it appears you want to solve by completing the square
x^2 + b/a x + (b/2a)^2 = (x + b/2a)^2
So that gives
(x + b/2a)^2 = -c/a + b^2/4a2
(x + b/2a)^2 = (b^2-4ac)/(2a)^2
now take square roots
x + b/2a = ±√(b^2-4ac)/2a
x = (-b±√(b^2-4ac)) / 2a
which is the quadratic formula for solving the first equation
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.