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#