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

$a = - 2 , b = 16 , c = - 4 \to x = \frac{- b \pm \left(\sqrt{{b}^{2} - 4 a c}\right)}{2 a} = \frac{- 16 \pm \sqrt{{16}^{2} - 4 \left(- 2\right) \left(- 4\right)}}{2 \left(- 2\right)} = \frac{- 16 \pm \sqrt{224}}{-} 4 = \frac{- 16 \pm 4 \sqrt{14}}{-} 4 = 4 \pm - \sqrt{14}$