derivation of the quadratic formula\ndescribe how to derive the quadratic formula from a quadratic equation…

Explanation:

Step1: Start with standard form

The standard - form of a quadratic equation is $ax^{2}+bx + c=0$, where $a\neq0$. First, isolate the $x$ - terms: $ax^{2}+bx=-c$.

Step2: Make coefficient of $x^{2}$ equal to 1

Divide the entire equation by $a$: $x^{2}+\frac{b}{a}x=-\frac{c}{a}$.

Step3: Complete the square

Take half of the coefficient of $x$, square it and add it to both sides. The coefficient of $x$ is $\frac{b}{a}$, half of it is $\frac{b}{2a}$, and its square is $\frac{b^{2}}{4a^{2}}$. So, $x^{2}+\frac{b}{a}x+\frac{b^{2}}{4a^{2}}=-\frac{c}{a}+\frac{b^{2}}{4a^{2}}$.

Step4: Rewrite the left - hand side as a perfect square

The left - hand side is $(x + \frac{b}{2a})^{2}$, and the right - hand side is $\frac{b^{2}-4ac}{4a^{2}}$ after getting a common denominator. So, $(x+\frac{b}{2a})^{2}=\frac{b^{2}-4ac}{4a^{2}}$.

Step5: Take the square root of both sides

$x+\frac{b}{2a}=\pm\frac{\sqrt{b^{2}-4ac}}{2a}$.

Step6: Solve for $x$

Subtract $\frac{b}{2a}$ from both sides: $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.

Answer:

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