Solve the inequality: -1/4(x^2-4x+3)> 0

Ok so assume that if you have
xy=0 then x and y=0
-1/4(x^2-4x+3)>0
multiply both sides by -4 to clear fraction (-4/-4=1)
flip sign
x^2-4x+3<0
factor
(x-3)(x-1)<0

we know that (+) times (-)=(-) and 
(-) times (-)=(+) so we don't want them to be both negative, we want different sign
we cannnot have 3 since it would be (0)(2)<0 which is false
we cannot have 1 either since (-2)(0)<0 is also false
lets see if the solution is in betwen 3 and 1 or outside of 3 and 1
ltry 2
2^3-4(2)+3<0
8-8+3<0
3<0
false
therefor it is outside ie
x>3 and x<1

5\16 x 2\5 = 10/80simplify to 1/80

1.) 18+5 is one way i’m pretty sure there are others but i can’t think of them right now