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

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Answer 1 · 2018-07-19T05:04:16

2 complex roots:
$x= (3+-isqrt(55))/(16)$

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

$x= (-6+-sqrt(6^2-4(-16)(-4)))/(2(-16))$

$x= (-6+-sqrt(36-256))/(-32)$

$x= (-6+-sqrt(-220))/(-32)$

$x= (-6+-2isqrt(55))/(-32)$

No real roots

2 complex roots
$x= (3+-isqrt(55))/(16)$

Answer 2 · 2018-07-19T05:12:59

2 imaginary roots: $t = 3/16 +- (sqrt(55))/16 i$

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

The quadratic formula can be used when the equation is in the form: $At^2 + Bt + C = 0$

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

$t = (-6 +- sqrt(36 - 4(-16)(-4)))/(2(-16)) = -6/-32 +- sqrt(-220)/-32$

$t = 3/16 +- (sqrt(4)sqrt(55)sqrt(-1))/-32$

$t = 3/16 +- (2sqrt(55)i)/-32$

$t = 3/16 +- (sqrt(55))/-16 i$

$t = 3/16 +- (sqrt(55))/16 i$

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