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 #2
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The quadratic formula is as follows:

When the quadratic function equals 0

 

 $${\mathtt{0}} = {{\mathtt{ax}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{bx}}{\mathtt{\,\small\textbf+\,}}{\mathtt{c}}$$

 

For example in:

 

$${\mathtt{0}} = {{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}$$ (a=1, b=2 and c=-3)

 

For these functions you can use the quadratic formula to find the value of x.

The formula being:

 

$${\mathtt{x}} = {\frac{\left({\mathtt{\,-\,}}{\mathtt{b}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{{\mathtt{b}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{a}}{\mathtt{\,\times\,}}{\mathtt{C}}}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{a}}\right)}}$$

 

Now you just fill in the values of a, b and c.

For the example I've used it would be:

 

$${\mathtt{x}} = {\frac{\left({\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{{\mathtt{2}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{1}}{\mathtt{\,\times\,}}-{\mathtt{3}}}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{1}}\right)}}$$

 

$${\mathtt{x}} = {\frac{\left({\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{4}}{\mathtt{\,\small\textbf+\,}}{\mathtt{12}}}}\right)}{{\mathtt{2}}}}$$

 

$${\mathtt{x}} = {\frac{\left({\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{16}}}}\right)}{{\mathtt{2}}}}$$

 

Now since the square root of any number (positive of course) can be both positive and negative, we get two solutions: The square root of 16 can be either 4 or -4, since negative negative becomes positive.

The solution would be either

 

$${\mathtt{x}} = {\frac{\left({\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}\right)}{{\mathtt{2}}}} = {\frac{{\mathtt{2}}}{{\mathtt{2}}}} = {\mathtt{1}}$$

or

$${\mathtt{x}} = {\frac{\left({\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,-\,}}{\mathtt{4}}\right)}{{\mathtt{2}}}} = {\frac{\left(-{\mathtt{6}}\right)}{{\mathtt{2}}}} = -{\mathtt{3}}$$

Squaddy May 7, 2015