$$x^{ 2 }-2x-15=0$$

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?

• Denominator should be $2$, not $-2$
– MPW
Commented Aug 23, 2014 at 12:25
• @Cherry_Developer, just fyi: \pm in math mode gives you a nice-looking $\pm$. Commented Aug 23, 2014 at 12:27

$$ax^2+bx+c=0\implies x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
$$\implies x^2-2x-15=0\implies x=\frac{2\pm\sqrt{(-2)^2-4\cdot1(-15)}}{2\cdot1}=\frac{2\pm8}2=5,-3$$
• That was a typo. I mean to put $$(x-5) (x+3)$$ Commented Aug 23, 2014 at 12:22
• Now I see the mistake I made. I was thinking $$a*b$$ goes in the denominator for some odd reason. It's supposed to be $$2a$$. got it. Commented Aug 23, 2014 at 12:28
Your error is that the denominator should be $2$, not $-2$
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 $$x=5$$ $$x=-3$$