How do you solve the quadratic using the quadratic formula given #3t^2+8t+5=-2t^2# over the set of complex numbers?

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Answer 1 · 2016-12-24T08:04:45

The solutions are #S={-4/5+3/5i,-4/5-3/5i}#

We need

#Delta=b^2-4ac#

and #x=(-b+-sqrtDelta)/(2a)#

Let's rewrite the equation

#3t^2+8t+5=-2t^2#

#5t^2+8t+5=0#

#Delta=8^2-4*5*5=64-100=-36#

As, #Delta<0#, the solutions are in #CC#

#t=(-8+-6i)/10#

#t=-4/5+-3/5i#