Question

Solve for t. d=−16t2+4t t=12±41−4d‾‾‾‾‾‾√ t=8±1−4d‾‾‾‾‾‾√ t=18±1−4d√8 t=12±41−4d√2

Answer 1

`t=(1)/(8)\left(1\pm√(1-4d)\right)`

Explanation:

Rearranging your quadratic to standard form, you get ...

16t^2 -4t +d = 0

For the quadratic formula, you have ...

a = 16

b = -4

c = d

so the solution is ...

`t=(-b\pm√(b^2-4ac))/(2a)\\\\t=(-(-4)\pm√((-4)^2-4(16)(d)))/(2(16))=(4\pm 4√(1-4d))/(32)\\\\t=(1)/(8)\left(1\pm√(1-4d)\right)`

Answer 2

C.
`t=(1)/(8)\pm(√((1-4\cdot d)))/(8)`

Explanation:

We have been given an equation
`d=-16t^2+4t`. We are asked to solve for t.

To solve for t we will rewrite our given equation in general form of equation
`(ax^2+bx+c=0)`.

`d+16t^2-4t=0`

`16t^2-4t+d=0`

Since our given equation is quadratic, so we will use quadratic formula to solve for t.

`x=(-b\pm√(b^2-4ac))/(2a)`

Upon comparing our given equation with general form of equation we can see,

`a=16`

`b=-4`

`c=d`

Upon substituting our given values in quadratic formula we will get,

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

`t=(4\pm√(16-64\cdot d))/(32)`

Upon factoring out 16 inside the square root we will get,

`t=(4\pm√(16(1-4\cdot d)))/(32)`

`t=(4\pm 4√((1-4\cdot d)))/(32)`

`t=(4)/(32)\pm(4√((1-4\cdot d)))/(32)`

`t=(1)/(8)\pm(√((1-4\cdot d)))/(8)`

Therefore, the solutions of t are
`t=(1)/(8)\pm(√((1-4\cdot d)))/(8)` and option C is the correct choice.