Abasket ball is passed through the air. the height h of the ball in feet after the distance d in feet the ball travels horizontally is given by h=-d^2+10d+5. how far horizontally from the player passing the ball will the ball land on the ground

The distance traveled will be the value of d when h=0 so:

-d^2+10d+5=0  which is the same as:

d^2-10d-5=0  now complete the square to find d, add 5 to both sides

d^2-10d=5  halve the linear coefficient, square it, and add it to both sides

d^2-10d+25=30  now the left side is a perfect square...

(d-5)^2=30  take the square root of both sides

d-5=±√30  add 5 to both sides

d=5±√30 since d>0

d=5+√30  ft

d≈10.48 ft (to nearest hundredth of a foot)

Idon't know that language