solve the equation

Question

16y(2–y)+(4y–5)^2=0

16y(2–y)+(4y–5)²=0

Answer 1

16y(2–y)+(4y–5)²=0

 

Begin by expanding: 32y -16y² = 16y² -40y + 25

 

Then go through the usual rigamarole of combining like terms and solving the quadratic equation. In the worst case, use the quadratic formula.

Answer 2

16y(2-y) +(4y-5)² = 0

 

32y - 16y² + 16y² - 40y + 25 = 0

 

-8y + 25 = 0

 

-8y = -25

 

  y = 25/8