quadratic formula derive $ax^2+bx+c$

\begin{align*} ax^{2} + bx + c = 0 & \Longleftrightarrow 4a^{2}x^{2} + 4abx + 4ac = 0\\\\ & \Longleftrightarrow (4a^{2}x^{2} + 4abx + b^{2}) = b^{2} - 4ac \tag{2}\\\\ & \Longleftrightarrow (2ax + b)^{2} = b^{2} - 4ac\\\\ & \Longleftrightarrow 2ax + b = \pm\sqrt{b^{2} - 4ac}\\\\ & \Longleftrightarrow x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \end{align*}

how does in step-2 4abx simplified to b. if it is divided by 4ax then it should be on both side. step-2 can someone explain it in detail?

• In step two: 1 transfer $4ac$ to RHS and add $b^2$ to both sides. 2 Express LHS as a square. Commented Aug 6, 2022 at 7:04
• Hint: If you expand $(2ax + b)^2$, then you get $$(2ax + b)^2 = (2ax)^2 + 2(2ax)b + b^2 = 4a^2 x^2 + 4abx + b^2.$$ @davidritchie: Is this what you meant to ask about? Commented Aug 6, 2022 at 7:32
• Please read the answer provided by Andre Nicolas to the question Why can ALL quadratic equations be solved by the quadratic formula?. Commented Aug 6, 2022 at 8:36

The $$b$$ comes from equating the middle term in the equation: $$(2ax + \triangle)^2 = 4a^2x^2 + 4abx + b^2$$. So the missing $$\triangle$$ is the $$b$$ you are talking about. It's a standard completing square technique in an intermediate algebra course at the college level.
$$ax^2+bx+c=0\\$$ Assuming $$a\neq 0$$, divide by $$a$$ and complete the square: $$(x+b/2a)^2-b^2/4a^2+c/a=0\\$$ Move everything without $$x$$ to the right and take the square root: $$x+b/2a=\pm\sqrt{{b^2 \over 4a^2}-{c\over a}}\\={\pm 1\over 2a}\sqrt{b^2-4ac}\\$$ Subtract $$b/2a$$ from both sides $$x={-b\pm\sqrt{b^2-4ac}\over 2a}\\$$