An archway is modeled by the equation y = -2x2 + 8x. a rod is to be placed across the archway at an angle defined by the equation x − 2.23y + 10.34 = 0. if the rod is attached to the archway at points a and b, such that point b is at a higher level than point a, at what distance from the ground level is point b?

1) y= - 2x² + 8x. It's a parabola open downward (a<0)  
2) x - 2.23.y + 10.34 = 0 . Re-write it : y = (x/2.23) + (10.34/2.23), a linear equation.
To find the intersections between 1) & 2), let 1) = 2)
-2x² + 8x = (x/2.23) + (10.34/2.23) 
-2x² + 8x - (x/2.23) - (10.34/2.23) =0 ; solve this quadratic for x values:

x' (that is A) = 0.772 & x" (that is B) = 3. (these are the values of x-intercept (parabola with line). To calculate the y-values, plug x' & x' in the equation:
for x' = 0.772, y = 0.34 → B(0.772 , 0.34)
for x" = 3, y = 0.016     → A(3 , 0.O16)
So B IS AT 0.34 Unit from the ground