# How do I solve the formula 16t^2 - vt - 40 = 0 for t?

Jun 16, 2018

$t = \frac{v \pm \sqrt{{v}^{2} + 2560}}{32}$

#### Explanation:

We have:

$16 {t}^{2} - v t - 40 = 0$

We can complete the square, by first dividing by the coefficient of ${t}^{2}$

$\therefore 16 \left\{{t}^{2} - \frac{v}{16} t - \frac{40}{16}\right\} = 0$

$\therefore {t}^{2} - \frac{v}{16} t - \frac{5}{2} = 0$

Then we factor half the coefficient of $t$:

$\therefore {\left(t - \frac{v}{32}\right)}^{2} - {\left(\frac{v}{32}\right)}^{2} - \frac{5}{2} = 0$

Which we can now readily rearrange for $t$:

$\therefore {\left(t - \frac{v}{32}\right)}^{2} = {v}^{2} / 1024 + \frac{5}{2}$

$\therefore {\left(t - \frac{v}{32}\right)}^{2} = {v}^{2} / 1024 + \frac{2560}{1024}$

$\therefore {\left(t - \frac{v}{32}\right)}^{2} = \frac{{v}^{2} + 2560}{1024}$

$\therefore t - \frac{v}{32} = \pm \sqrt{\frac{{v}^{2} + 2560}{1024}}$

$\therefore t - \frac{v}{32} = \pm \frac{\sqrt{{v}^{2} + 2560}}{32}$

$\therefore t = \frac{v}{32} \pm \frac{\sqrt{{v}^{2} + 2560}}{32}$

$\therefore t = \frac{v \pm \sqrt{{v}^{2} + 2560}}{32}$