How do you determine the number of real solutions to -3x^2+4x+1=0?

1 Answer
Oct 30, 2014

-3x^2+4x+1=0
rArr 3x^2-4x-1=0
a=3, b=-4, c=-1

D=b^2-4ac
rArr D= (-4)^2 - 4 (3)(-1)
rArr D= 16 + 12
rArr D = 28

As Discriminant is greater than zero, there are two real solutions.

PS :
If D < 0, there are no real solutions
If D = 0, there is only one real solution