How do you solve the quadratic with complex numbers given #-6x^2+12x-7=0#?

Question

2519 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-06T04:45:09

#x=(1+2sqrt(6i))/6, (1-2sqrt(6i))/6#

Solve:

#-6x^2+12x-7=0# is a quadratic equation in standard form: #ax^2+bx+c#, where #a=-6#, #b=12#, and #c=-7#.

The quadratic formula can be used to solve this equation.

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

Substitute the given values into the formula.

#x=(-12+-sqrt(12^2-4xx-6xx-7))/(2xx-6)#

#x=(-12+-sqrt(144-168))/(-12)#

#x=(-12+-sqrt(-24))/(-12)#

Factor #-24#.

#x=(-12+-sqrt(2xx2xx2xx3i))/(-12)#

#x=(-12+-2sqrt(6i))/(-12)#

Simplify by dividing by #-12#.

#x=(1+-2sqrt(6i))/6#

Solutions for #x#.

#x=(1+2sqrt(6i))/6, (1-2sqrt(6i))/6#