Question

Solve for t. use the quadratic formula. d=−16t^2+12t

Answer 1

Answer:

$t\,=\,\frac{-3+\sqrt{9+4d}}{-8}\:\:and\:\:\frac{3+\sqrt{9+4d}}{8}$

Step-by-step 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

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

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,

$t\,=\,\frac{-12\pm\sqrt{12^2-4\times(-16)\times d}}{2\times(-16)}$

$t\,=\,\frac{-12\pm\sqrt{144+64d}}{-32}$

$t\,=\,\frac{-12\pm\sqrt{16(9+4d)}}{-32}$

$t\,=\,\frac{-12\pm4\sqrt{9+4d}}{-32}$

$t\,=\,\frac{4(-3\pm\sqrt{9+4d})}{-32}$

$t\,=\,\frac{-3\pm\sqrt{9+4d}}{-8}$

$t\,=\,\frac{-3+\sqrt{9+4d}}{-8}\:\:,\:\:\frac{-3-\sqrt{9+4d}}{-8}$

$t\,=\,\frac{-3+\sqrt{9+4d}}{-8}\:\:and\:\:\frac{-(3+\sqrt{9+4d})}{-8}$

$t\,=\,\frac{-3+\sqrt{9+4d}}{-8}\:\:and\:\:\frac{3+\sqrt{9+4d}}{8}$

Therefore, $t\,=\,\frac{-3+\sqrt{9+4d}}{-8}\:\:and\:\:\frac{3+\sqrt{9+4d}}{8}$

Answer 2

Answer:

$\frac{3-\sqrt{9-d}} {8}\text{ or }t=\frac{3+\sqrt{9-d}} {8}$

Step-by-step 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,

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

By comparing equation (1) and (2),

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

$t=\frac{-12\pm \sqrt{12^2-4\times -16\times -d}}{2\times -16}$

$t = \frac{-12\pm \sqrt{16\times 9-16\times d}}{-32}$

$t = \frac{-12\pm \sqrt{16}\times \sqrt{9-d}} {-32}$

$t = \frac{-12\pm 4\sqrt{9-d}} {-32}$

$t = \frac{4(-3\pm \sqrt{9-d})} {4(-8)}$

$t = \frac{-3\pm \sqrt{9-d}} {-8}$

$t = \frac{-3+\sqrt{9-d}} {-8}\text{ or }t=\frac{-3-\sqrt{9-d}} {-8}$

$\implies t = \frac{3-\sqrt{9-d}} {8}\text{ or }t=\frac{3+\sqrt{9-d}} {8}$

Which is the required solution.