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

Answer:

$t=\dfrac{1}{8}\left(1\pm\sqrt{1-4d}\right)$

Step-by-step 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 ...

$t=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\t=\dfrac{-(-4)\pm\sqrt{(-4)^2-4(16)(d)}}{2(16)}=\dfrac{4\pm 4\sqrt{1-4d}}{32}\\\\t=\dfrac{1}{8}\left(1\pm\sqrt{1-4d}\right)$

Answer 2

Answer:

C. $t=\frac{1}{8}\pm\frac{\sqrt{(1-4\cdot d)}}{8}$

Step-by-step 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.

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

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,

$t=\frac{--4\pm\sqrt{(-4)^2-4\cdot 16\cdot d}}{2\cdot 16}$

$t=\frac{4\pm\sqrt{16-64\cdot d}}{32}$

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

$t=\frac{4\pm\sqrt{16(1-4\cdot d)}}{32}$

$t=\frac{4\pm 4\sqrt{(1-4\cdot d)}}{32}$

$t=\frac{4}{32}\pm\frac{4\sqrt{(1-4\cdot d)}}{32}$

$t=\frac{1}{8}\pm\frac{\sqrt{(1-4\cdot d)}}{8}$

Therefore, the solutions of t are$t=\frac{1}{8}\pm\frac{\sqrt{(1-4\cdot d)}}{8}$ and option C is the correct choice.