The roots are fairly simple to find, no need to invoke quadratic formula. First we factor,
[tex]x^2-15x-16=0\implies (x-16)(x+1)=0[/tex]
Then we get zeros at [tex]x_1=16,x_2=-1[/tex].
Using veita's formulas we can check it namely,
[tex]
x_1x_2=c/a\implies 16(-1)=-16/1\implies -16=-16\\
x_1+x_2=-b/a\implies 16-1=-(-15/1)\implies 15=15
[/tex]
So as you can see solutions pass both tests and therefore are valid.
Hope this helps.
