How do you solve #x^2 + 3x + 6 = 0# by quadratic formula?

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Answer 1 · 2015-12-30T12:04:55

Substitute the coefficients #a=1#, #b=3# and #c=6# into the quadratic formula to find:

#x=(-3+-i sqrt(15))/2#

#x^2+3x+6# is in the form #ax^2+bx+c# with #a=1#, #b=3# and #c=6#.

First note that the discriminant #Delta# is negative.

#Delta = b^2-4ac = 3^2-(4xx1xx6) = 9 - 24 = -15#

So our quadratic equation has two Complex roots.

The roots are given by the quadratic formula:

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

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

#=(-3+-sqrt(-15))/2#

#=(-3+-i sqrt(15))/2#

Answer 2 · 2015-12-30T12:07:05

The solutions are:
#x= color(blue)((-3+sqrt(-15))/2 #

#x= color(blue)((-3-sqrt(-15))/2 #

#x^2 +3x +6 =0#

The equation is of the form #color(blue)(ax^2+bx+c=0# where:
#a=1, b=3, c=6#

The Discriminant is given by:
#Delta=b^2-4*a*c#

# = (3)^2-(4*(1)*6)#

# = 9-24 = -15#

As #Delta = -15#, this equation has NO REAL SOLUTIONS

The solutions are found using the formula:
#x=(-b+-sqrtDelta)/(2*a)#

#x = ((-3)+-sqrt(-15))/(2*1) = (-3+-sqrt(-15))/2#

The solutions are:
#x= color(blue)((-3+sqrt(-15))/2 #

#x= color(blue)((-3-sqrt(-15))/2 #