Question 259972
<pre><font size = 4 color = "indigo"><b>
I think you aren't being careful with your signs and
with putting parentheses around what you are substituting.
If your book says the solutions are -5 and -3, then you 
have typed one or two signs wrong in the original problem, 
or else the book has a typographical error, and sometimes 
they do.

You gave the problem as

{{{x^2-8x-15=0}}}

but the solutions to that are not -5 and -3.

You would have to have had this problem instead

{{{x^2+8x+15=0}}}  [not {{{x^2-8x+15=0}}} as the other tutor said.]

In order to get solutions -5 and -3.

Then you would factor and get

{{{(x+5)(x+3)=0}}}

and then you'd set each of those = 0 and get

x+5=0      x+3=0
  x=-5       x=-3


But let's suppose it was as you thought

You have this:

{{{x^2-8x-15=0}}}
x={{{(-(-8)+- sqrt((-8)^2-4*1*(-15)))/(2*1)}}}
x={{{8+- sqrt (64+60)/(2)}}}
x={{{8+- sqrt (124)/(2)}}}
x={{{8+- sqrt (4*31)/(2)}}}
x={{{(8+- 2sqrt(31))/(2)}}}

Then you'd factor 2 out of the top:

x={{{(2(4+- sqrt(31)))/(2)}}}

and then cancel the 2's

x={{{(cross(2)(4+- sqrt(31)))/(cross(2))}}}

{{{x=4+- sqrt(31)}}}

Edwin</pre>