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# Answer on Algebra Question for tommy

Question #2929
How can calculate the Quadratic Equation, for example, 3x2+6x-3

&lt;img src=&quot;/cgi-bin/mimetex.cgi?ax%5E2%20+%20bx%20+%20c%20=0&quot; title=&quot;ax^2 + bx + c =0&quot;&gt;
the answer can be found by the formula:
&lt;img src=&quot;http://latex.codecogs.com/gif.latex?x%20=%20%5Cfrac%20%7B-b%20%5Cpm%20%5Csqrt%20%7Bb%5E2-4ac%7D%7D%7B2a%7D&quot; title=&quot;x = \frac {-b \pm \sqrt {b^2-4ac}}{2a}&quot;&gt;, when &lt;img src=&quot;http://latex.codecogs.com/gif.latex?b%5E2-4ac%20%5Cge%200&quot; title=&quot;b^2-4ac \ge 0&quot;&gt;. If this expression is negative, there is no any roots, of they are complex.
For example:
&lt;img src=&quot;http://latex.codecogs.com/gif.latex?3x%5E2+6x%20-3%20=0%20%5C%5C%20x%20=%20%5Cfrac%7B-6%20%5Cpm%20%5Csqrt%7B6%5E2%20-%204%5Ccdot3%5Ccdot3%7D%7D%7B2%5Ccdot3%7D%20=%20%5Cfrac%7B-6%20%5Cpm%200%7D%7B6%7D%20=%20-1&quot; title=&quot;3x^2+6x -3 =0 \\ x = \frac{-6 \pm \sqrt{6^2 - 4\cdot3\cdot3}}{2\cdot3} = \frac{-6 \pm 0}{6} = -1&quot;&gt; - only one root.

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