Determine the type and number of solutions of 3x^2+6x+4=0
I hope this helps you
3x^2+6x+4=0
a=3 b=6 c=4
disctirminant=b^2-4ac
disctirminant=6^2-4.3.4
disctirminant=36-48
disctirminant= -12
x1= -6+ square root of -12/2.3
x1= -6+2i square root of 3/6
x1= -3+i square root of 3/3
x2= -6-2i square root of 3/6
x2= -3-I square root of 3/3
Step 1. Calculate the discriminant:
\(\Delta = b^2 - 4ac\)
Step 2.
If \(\Delta>0\), there are two solutions, both real numbers
If \(\Delta=0\), there is only one solution, real number
If \(\Delta<0\), there are two solutions, both complex numbers
In your case:
a=3, b=6, c=4
\(\Delta = 6^2 - 4 \cdot 3 \cdot 4 = -12\)
So you have two solutions, both complex numbers