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

Jun 5, 2017

color(blue)(h=-8+2sqrt17 or h=-8-2sqrt17

#### Explanation:

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

$h = - \left(- {t}^{2}\right) + 16 t - 4$

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

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

color(blue)(a^2+bx+c

color(blue)(a=1,b=16,c=-4

$h = \frac{- \left(16\right) \pm \sqrt{{\left(16\right)}^{2} - 4 \left(1\right) \left(- 4\right)}}{2 \left(1\right)}$

$\frac{- 16 \pm \sqrt{256 + 16}}{2}$

$\frac{- 16 \pm \sqrt{272}}{2}$

$\frac{- 16 \pm \sqrt{2 \cdot 2 \cdot 2 \cdot 2 \cdot 17}}{2}$

color(blue)(sqrt2 xx sqrt2=2

$\frac{- 16 \pm 4 \sqrt{17}}{2}$

$\frac{{\cancel{4}}^{\textcolor{b l u e}{2}} \left(- 4 + \sqrt{17}\right)}{\cancel{\textcolor{b l u e}{2}}} ^ \textcolor{b l u e}{1} \mathmr{and} \frac{{\cancel{4}}^{\textcolor{b l u e}{2}} \left(- 4 - \sqrt{17}\right)}{\cancel{2}} ^ \textcolor{b l u e}{1}$

2(-4+sqrt17) or 2(-4-sqrt17))

color(blue)(-8+2sqrt17 or -8-2sqrt17