Here you have quadratic equation which has parabola for it's graph. So in order to find how high will the ball reach you actually need to find coordinates of the vertex of parabola.

Coordinates of vertex for `y(x)=ax^2+bx+c`

`V(-b/(2a),(4ac-b^2)/(4a))`

The first coordinate tells us the distance and the second coordinate tells us the height which is what we want to know.

`h=(4cdot(-16)cdot6-80^2)/(4cdot(-16))=(-384-6400)/(-64)=106`

The ball will reach 106.

When will the ball land?

The ball is on the land when it's height is equal to 0. Hence we have equation

`-16x^2+80x+6=0`

We can solve that by using quadratic formula `x_(1,2)=(-b pm sqrt(b^2-4ac))/(2a)`

`x_(1,2)=(-80pm sqrt(6400+384))/(-32)approx(-80pm82.37)/(-32)`

`x_1approx5.07`

`x_2approx-0.7` <-- This is 0.7s before launch so it's not solution to our problem.

The ball will land 5.07 seconds after launch.