it is a quadratic
where x-8 is the squared term
do
x-8=u
u^2-13u+30=0
solve
u=[tex] \frac{-b+/- \sqrt{b^2-4ac} }{2a} [/tex]
u=[tex] \frac{-(-13)+/- \sqrt{(-13)^2-4(1)(30)} }{2(1)} [/tex]
u=[tex] \frac{13+/- \sqrt{169-120} }{2} [/tex]
u=[tex] \frac{13+/- \sqrt{49} }{2} [/tex]
u=[tex] \frac{13+/- 7 }{2} [/tex]
u=[tex] \frac{20} }{2} [/tex] or u=[tex] \frac{6 }{2} [/tex]
u=10 or 3
10=x-8 or 3=x-8
18=x or 11=x
x=11 and/or 18
