# Solving a simple quadratic equation

I have problems every time I face a quadratic equation. What can I do to learn how to solve them? Can anyone please show me how to solve the one below and explain the basic principle of solving quadratic equations.

$$x^2- xa - ab = 0$$

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There is a formula: $$Ax^2 + Bx + C = 0 \quad \Rightarrow \quad x_{1,2} = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}.$$ $A$ is whatever is next to $x^2$, $B$ is whatever is next to $x$, and $C$ is without $x$. In your case: $$x^2 - xa - ab = 1 \cdot x^2 + (-a)x + (-ab) = 0 \quad \Rightarrow \quad A = 1, \quad B = -a, \quad C = -ab,$$ so $$x_{1,2} = \frac{a \pm \sqrt{a^2 + 4ab}}{2}.$$

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Where does this formula come from? – 71GA Jul 15 '13 at 15:09
It's a well known formula which you can easily check by substituting $x_1$ and $x_2$ for $x$. @JamesMaslek has just give the proper link, so I won't repeat what he wrote. – Vedran Šego Jul 15 '13 at 15:15
the formula is actually derived in a different question on this site – Zar Jul 15 '13 at 15:17
Here is the question @Zar is talking about: math.stackexchange.com/questions/49229/… – EsX_Raptor Jul 15 '13 at 15:37

$x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$ (as @Vedran Sego has). This comes from completing the square