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