How do you solve #3z^2+10z+15=0#?

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Answer 1 · 2015-06-15T23:07:54

The discriminant is negative, so I would use the quadratic formula directly to get:

#z = (-5 +- 2sqrt(5) i)/3#

#3z^2+10z+15# is of the form #az^2+bz+c# with #a=3#, #b=10# and #c=15#.

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = 10^2 - (4xx3xx15) = 100 - 180 = -80#

Since this is negative the quadratic equation has two distinct complex roots.

The roots are given by the quadratic formula:

#z = (-b+-sqrt(Delta))/(2a) = (-10+-sqrt(-80))/(2*3)#

#=(-10+-sqrt(80)i)/6#

#=(-10+-sqrt(16*5)i)/6#

#=(-10+-4sqrt(5)i)/6#

#=(-5+-2sqrt(5)i)/3#