How do you solve the quadratic using the quadratic formula given #7-8z^2=6z+16# over the set of complex numbers?

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Answer 1 · 2016-11-20T09:08:18

The solution is #S={(-3+3isqrt7)/(8),(-3-3isqrt7)/(8)} #

Rearranging the equation

#8z^2+6z+9=0#

Let's calculate the discriminant

#Delta=b^2-4ac=36-4*8*9=36-288=-252#

#Delta<0#, so solutions are in #CC#

Then,
#z=(-b+-sqrtDelta)/(2a)=(-6+-isqrt252)/16#

#=(-6+-6isqrt7)/16=(-3+-3isqrt7)/(8)#