How do you solve the quadratic using the quadratic formula given #3a^2-4a-4=0# over the set of complex numbers?

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Answer 1 · 2016-08-15T08:33:13

This equation has 2 real roots: #a_1=-2/3# and #a_2=2#

To solve this equation using the quadratic formula first we calculate the determinant:

#Delta=b^2-4ac #

#Delta=(-4)^2-4*3*(-4)=16+48=64#

#Delta>0# so the equation has 2 real roots:

#x_1=(-b-sqrt(Delta))/(2a)# and #x_2=(-b+sqrt(Delta))/(2a)#

#a_1=(4-8)/6=-2/3#

#a_2=(4+8)/6=2#

Now we can write the answer:
The equation has 2 solutions #a_1=-2/3# and #a_2=2#.