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

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Answer 1 · 2016-03-05T20:20:09

$a=-2,b=16,c=-4 ->x=(-b+- (sqrt(b^2-4ac)))/(2a) = (-16+- sqrt(16^2-4(-2)(-4)))/(2(-2)) = (-16+-sqrt224)/-4 = (-16+-4sqrt14)/-4 = 4+- -sqrt14$

List a,b,c and then substitute in to the quadratic formula then solve for x

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