Question

How do you solve `4y ^ { 2} + 19= 20y`?

Answer 1

`y = (5-sqrt6)/2 " " and " " y = (5+sqrt6)/2`

Explanation:

First, subtract `20y` from both sides to get:

`4y^2 - 20y + 19 = 0`

Now, we can use the quadratic formula to solve for `y`.

`a = 4`
`b = -20`
`c = 19`

`y = (-b+-sqrt(b^2-4ac))/(2a)`

Plugging in these values, we get:

`y = (-(-20)+-sqrt((-20)^2-4(4)(19)))/(2(4))`

`y = (20+-sqrt(400 - 304))/8`

`y = (20+-sqrt96)/8`

`y = (20+-sqrt(16 * 6))/8`

`y = (20+-4sqrt6)/8`

`y = (5+-sqrt6)/2`

Therefore, these are our two solutions:

`y = (5-sqrt6)/2 " " and " " y = (5+sqrt6)/2`

Final Answer