Question

Solve for t. use the quadratic formula.

d=−16t^2+12t

Answer 1

`\frac{3-√(9-d)} {8}\text{ or }t=\frac{3+√(9-d)} {8}`

Explanation:

Here, the given expression,

`d= -16t^2+12t`

`\implies -16x^2+12t-d=0` ------(1)

Since, if a quadratic equation is,

`ax^2+bx+c=0` ------(2)

By using quadratic formula,

We can write,

`x=(-b\pm √(b^2-4ac))/(2a)`

By comparing equation (1) and (2),

We get, a = -16, b = 12, c = -d,

`t=(-12\pm √(12^2-4* -16* -d))/(2* -16)`

`t = (-12\pm √(16* 9-16* d))/(-32)`

`t = \frac{-12\pm √(16)* √(9-d)} {-32}`

`t = \frac{-12\pm 4√(9-d)} {-32}`

`t = \frac{4(-3\pm √(9-d))} {4(-8)}`

`t = \frac{-3\pm √(9-d)} {-8}`

`t = \frac{-3+√(9-d)} {-8}\text{ or }t=\frac{-3-√(9-d)} {-8}`

`\implies t = \frac{3-√(9-d)} {8}\text{ or }t=\frac{3+√(9-d)} {8}`

Which is the required solution.

Answer 2

`t\,=\,(-3+√(9+4d))/(-8)\:\:and\:\:(3+√(9+4d))/(8)`

Explanation:

Given: d = -16t² + 12t

To find: t using quadratic formula

If we have quadratic equation in form ax² + bx + c = 0

then, by quadratic formula we have

`x\,=\,(-b\pm√(b^2-4ac))/(2a)`

Rewrite the given equation,

-16t² + 12t - d = 0

from this equation we have,

a = -16 , b = 12 , c = d

now using quadratic formula we get,

`t\,=\,(-12\pm√(12^2-4*(-16)* d))/(2*(-16))`

`t\,=\,(-12\pm√(144+64d))/(-32)`

`t\,=\,(-12\pm√(16(9+4d)))/(-32)`

`t\,=\,(-12\pm4√(9+4d))/(-32)`

`t\,=\,(4(-3\pm√(9+4d)))/(-32)`

`t\,=\,(-3\pm√(9+4d))/(-8)`

`t\,=\,(-3+√(9+4d))/(-8)\:\:,\:\:(-3-√(9+4d))/(-8)`

`t\,=\,(-3+√(9+4d))/(-8)\:\:and\:\:(-(3+√(9+4d)))/(-8)`

`t\,=\,(-3+√(9+4d))/(-8)\:\:and\:\:(3+√(9+4d))/(8)`

Therefore,
`t\,=\,(-3+√(9+4d))/(-8)\:\:and\:\:(3+√(9+4d))/(8)`