# How do you find the roots, real and imaginary, of  h=-16t^2+6t -4  using the quadratic formula?

Jul 19, 2018

2 complex roots:
$x = \frac{3 \pm i \sqrt{55}}{16}$

#### Explanation:

x=(-B+-sqrt(B^2-4AC))/(2A

$x = \frac{- 6 \pm \sqrt{{6}^{2} - 4 \left(- 16\right) \left(- 4\right)}}{2 \left(- 16\right)}$

$x = \frac{- 6 \pm \sqrt{36 - 256}}{- 32}$

$x = \frac{- 6 \pm \sqrt{- 220}}{- 32}$

$x = \frac{- 6 \pm 2 i \sqrt{55}}{- 32}$

No real roots

2 complex roots
$x = \frac{3 \pm i \sqrt{55}}{16}$

Jul 19, 2018

2 imaginary roots: $t = \frac{3}{16} \pm \frac{\sqrt{55}}{16} i$

#### Explanation:

Given: $h = - 16 {t}^{2} + 6 t - 4$

The quadratic formula can be used when the equation is in the form: $A {t}^{2} + B t + C = 0$

$t = \frac{- B \pm \sqrt{{B}^{2} - 4 A C}}{2 A}$

$t = \frac{- 6 \pm \sqrt{36 - 4 \left(- 16\right) \left(- 4\right)}}{2 \left(- 16\right)} = - \frac{6}{-} 32 \pm \frac{\sqrt{- 220}}{-} 32$

$t = \frac{3}{16} \pm \frac{\sqrt{4} \sqrt{55} \sqrt{- 1}}{-} 32$

$t = \frac{3}{16} \pm \frac{2 \sqrt{55} i}{-} 32$

$t = \frac{3}{16} \pm \frac{\sqrt{55}}{-} 16 i$

$t = \frac{3}{16} \pm \frac{\sqrt{55}}{16} i$