x(7-x)>8 perform indicated multiplication on left side
7x-x^2>8 divide both sides by -1
x^2-7x<-8 halve the linear coefficient, square it, and add it to both side ie (-7/2)^2=12.25
x^2-7x+12.25<4.25 now the left side is a perfect square
(x-3.5)^2<4.25 take the square root of both sides
x-3.5<±√4.25 add 3.5 to both sides
x<3.5±√4.25
So x>3.5-√4.25 and x<3.5+√4.25 or if you want approximations
x>1.44 and x<5.56 or in interval notation:
x=(1.44, 5.56) (to nearest hundredths)