How do you solve the quadratic with complex numbers given #5x^2+12x+8=0#?

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Answer 1 · 2016-12-25T08:13:10

The solutions are #S={-6/5-2/5i,-6/5+2/5i} #

We need #ax^2+bx+c=0#

#Delta=b^2-4ac#

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

The equation is #5x^2+12x+8=0#

#Delta=12^2-4*5*8=144-160=-16#

As, #Delta<0#, there are no roots in #RR# but in #CC#

#x=(-12+-4i)/10#

The roots are #x_1=-6/5-2/5i# and #x_2=-6/5+2/5i#