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

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Answer 1 · 2016-09-04T09:48:01

#x=-4/5-i3/5# and #x=-4/5+i3/5#

Quadratic formula gives the solution of quadratic equation #ax^2+bx+c=0# as

#x=(-b+-sqrt(b^2-4ac))/(2a)#

Hence solution of #5x^2+8x+5=0# is given by

#x=(-8+-sqrt(8^2-4×5×5))/(2×5)#

= #(-8+-sqrt(64-100))/10#

= #(-8+-sqrt(-36))/10#

= #(-8+-i6)/10=-4/5+-i3/5#

i.e. #x=-4/5-i3/5# and #x=-4/5+i3/5#