How do you solve #2x^2+6x+33=0#?

1 Answer
Oct 23, 2016

Answer:

#x = -3/2+-sqrt(57)/2i#

Explanation:

#2x^2+6x+33 = 0#

This is in the form:

#ax^2+bx+c = 0#

with #a = 2#, #b = 6# and #c = 33#.

The discriminant #Delta# is given by the formula:

#Delta = b^2-4ac = color(blue)(6)^2-4(color(blue)(2))(color(blue)(33)) = 36 - 264 = -228 = -2^2*57#

Since #Delta < 0# this quadratic has a Complex conjugate pair of roots, which we can find using the quadratic formula:

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

#color(white)(x) = (-b+-sqrt(Delta))/(2a)#

#color(white)(x) = (-6+-sqrt(-2^2*57))/(2*2)#

#color(white)(x) = (-6+-2sqrt(57)i)/4#

#color(white)(x) = -3/2+-sqrt(57)/2i#