Could you please help me and explain this issue:

If a quadratic equation is not easily factored then its roots can be found using quadratic formula: If $ax^2+bx+c=0$ ($a\ne0$), then the roots are $$x=\frac{-b+\sqrt{b^2-4ac}}{2a} \text{ and }x=\frac{-b-\sqrt{b^2-4ac}}{2a}$$ These are two distinct real numbers unless $b^2-4ac\le0$. If $b^2-4ac=0$ then these two expressions for $x$ are equal to $-\frac b{2a}$, and the equation has only one root. If $b^2-4ac<0$, then $\sqrt{b^2-4ac}$ is not a real number and the equation has no real roots.

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Is there any secyfic formula for this exist?


closed as unclear what you're asking by user91500, wythagoras, Martin Sleziak, Mark Bennet, Chappers Jun 21 '15 at 11:38

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see https://en.wikipedia.org/wiki/Quadratic_formula

take the quadratic equation ($ax^2 + bx + c = 0$)

Divide the quadratic equation by a, which is allowed because a is non-zero:

$$x^2 + \frac{b}{a} x + \frac{c}{a}=0.$$ Subtract $\frac c a$ from both sides of the equation, yielding:

$$x^2 + \frac{b}{a} x= -\frac{c}{a}.$$ The quadratic equation is now in a form to which the method of completing the square can be applied. Thus, add a constant to both sides of the equation such that the left hand side becomes a complete square:

$$x^2+\frac{b}{a}x+\left( \frac{b}{2a} \right)^2 =-\frac{c}{a}+\left( \frac{b}{2a} \right)^2,$$ which produces:

$$\left(x+\frac{b}{2a}\right)^2=-\frac{c}{a}+\frac{b^2}{4a^2}.$$ Accordingly, after rearranging the terms on the right hand side to have a common denominator, we obtain this:

$$\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}.$$ The square has thus been completed. Taking the square root of both sides yields the following equation:

$$x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac\ }}{2a}.$$ Isolating $x$ gives the quadratic formula:

$$x=\frac{-b\pm\sqrt{b^2-4ac\ }}{2a}.$$ The plus-minus symbol "$±$" indicates that both

$$x=\frac{-b + \sqrt {b^2-4ac}}{2a}\quad\text{and}\quad x=\frac{-b - \sqrt {b^2-4ac}}{2a}$$

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