Quadratic formula and factoring are leading to different answers $$x^{ 2 }-2x-15=0$$
By factoring, I get:
$$(x-5)(x+3)$$
Which has the solutions: 
$$x=5, x=-3$$
However when I use the quadratic formula (which is what the book saids to use), I get 
$$\frac { 2 \pm \sqrt { 4-(4\cdot1\cdot(-15)) }  }{ -2 } =$$
$$\frac { 2\pm 8 }{ -2 } $$
Which I evaluate to be $$x=-5, x=3$$
Where am I going wrong?

The quadratic formula is
$$
x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}
$$
The denominator should be $2$ because
\[ a=1 \]
\[ b=-2 \]
\[ c=-15 \]
So then
\[
x=\frac{2\pm \sqrt{4-4(-15)}}{2}= \frac{2\pm \sqrt{64}}{2}= \frac{2\pm 8}{2}=1\pm 4
\]
Thus
$$ x=5 $$
$$ x=-3 $$
