what is the values of x that are solutions to f(x)=0, where f(x)=3x^2+6x+7

Question

what is the value of x that are soutions to f(x)=3x^2+6x+7

Answer 1

Using the quadratic formula where A=3, B=6, C=7, x=(-B+/-sqrt(B^2-4AC))/2A=

-6+/-sqrt(36-4*3*7))/2*3

x=(-6+/-sqrt(36-84))/6

=(-6+/-sqrt(-48))/6

 

Note that since the square root is negative, neither solution is real.

 

Regards,

Deanna

Answer 2

f(x) = 3x² + 6x + 7

 

0 = 3x² + 6x + 7

Use the quadratic formula:

 

x = [-b±√(b²-4ac)]/2a

 

In this case, a=3, b=6, c=7

 

x = [-6±√(6²-4*3*7)]/6

 

x = [-6±√(-48)]/6                  [Roots will be complex]

 

x = -1 ± i2(√3)/3