Question

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

Answer 1

`-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