# How do I solve the formula 16t^2 - 12t - h = 0 for t?

May 26, 2018

$t = \frac{3 \pm \sqrt{4 h + 9}}{8}$

#### Explanation:

$16 {t}^{2} - 12 t - h = 0$

you would need to complete the square:

$a {x}^{2} + b x + c$

$a = 1$

$c = {\left(\frac{1}{2} b\right)}^{2}$

So first move the $h$ over:

$16 {t}^{2} - 12 t = h$

now divide by 16 so $a = 1$

$\frac{16 {t}^{2} - 12 t}{16} = \frac{h}{16}$

${t}^{2} - \frac{3}{4} t = \frac{h}{16}$

now we are ready to complete the square:

${t}^{2} - \frac{3}{4} t + c = \frac{h}{16} + c$

$c = {\left(\frac{1}{2} b\right)}^{2}$

$c = {\left(\frac{1}{2} \cdot - \frac{3}{4}\right)}^{2} = \frac{9}{64}$

${t}^{2} - \frac{3}{4} t + \frac{9}{64} = \frac{h}{16} + \frac{9}{64}$

${\left(t - \frac{3}{8}\right)}^{2} = \frac{h}{16} + \frac{9}{64}$

$\sqrt{{\left(t - \frac{3}{8}\right)}^{2}} = \pm \sqrt{\frac{h}{16} + \frac{9}{64}}$

$t - \frac{3}{8} = \pm \sqrt{\frac{h}{16} + \frac{9}{64}}$

$t = \frac{3}{8} \pm \sqrt{\frac{h}{16} + \frac{9}{64}}$

$t = \frac{3}{8} \pm \sqrt{\frac{4 h}{64} + \frac{9}{64}}$

$t = \frac{3}{8} \pm \sqrt{\frac{4 h + 9}{64}}$

$t = \frac{3}{8} \pm \frac{\sqrt{\left(4 h + 9\right)}}{8}$

$t = \frac{3 \pm \sqrt{4 h + 9}}{8}$