Question

Determine the type and number of solutions of 3x^2+6x+4=0

Answer 1

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 $\Delta$, 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

Answer 2

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