Question 593408
My problem is to use the quadratic equation for: (r-3)(r+9)=16.
FOIL
r^2 + 9r - 3r - 27 = 16
r^2 + 6x - 27 - 16 = 0
r^2 + 6x - 43 = 0
Use the quadratic formula
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
x=r, a=1, b=6, c=-43
{{{r = (-6 +- sqrt(6^2-4*1*-43 ))/(2*1) }}} 
:
{{{r = (-6 +- sqrt(36 + 172))/2 }}} 
:
{{{r = (-6 +- sqrt(208))/2 }}}
I think here you went astray somewhere
Factor inside the radical to reveal a perfect square
{{{r = (-6 +- sqrt(16*13))/2 }}}
Extract the square root of 16
{{{r = (-6 +- 4sqrt(13))/2 }}}
Two solutions
{{{r = (-6 + 4sqrt(13))/2 }}}
{{{r = -3 + 2sqrt(13)) }}}
and
{{{r = -3 - 2sqrt(13)) }}}