\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?
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 $$ x={-b\pm\sqrt{b^2-4ac}\over 2a}\\ $$
