# How do you use the quadratic formula to solve 16t^2-4t+3=0?

Sep 24, 2017

$t = \frac{1}{8} \pm \frac{1}{8} \sqrt{11} i$

#### Explanation:

Given:

$16 {t}^{2} - 4 t + 3 = 0$

This is in standard form for a quadratic equation:

$a {t}^{2} + b t + c = 0$

with $a = 16$, $b = - 4$ and $c = 3$

It has roots given by the quadratic formula:

$t = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$\textcolor{w h i t e}{t} = \frac{- \left(\textcolor{b l u e}{- 4}\right) \pm \sqrt{{\left(\textcolor{b l u e}{- 4}\right)}^{2} - 4 \left(\textcolor{b l u e}{16}\right) \left(\textcolor{b l u e}{3}\right)}}{2 \left(\textcolor{b l u e}{16}\right)}$

$\textcolor{w h i t e}{t} = \frac{4 \pm \sqrt{16 - 192}}{32}$

$\textcolor{w h i t e}{t} = \frac{4 \pm \sqrt{- 176}}{32}$

$\textcolor{w h i t e}{t} = \frac{4 \pm \sqrt{176} i}{32}$

$\textcolor{w h i t e}{t} = \frac{4 \pm \sqrt{16 \cdot 11} i}{32}$

$\textcolor{w h i t e}{t} = \frac{4 \pm 4 \sqrt{11} i}{32}$

$\textcolor{w h i t e}{t} = \frac{1}{8} \pm \frac{1}{8} \sqrt{11} i$