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

Oct 30, 2014

$- 3 {x}^{2} + 4 x + 1 = 0$
$\Rightarrow 3 {x}^{2} - 4 x - 1 = 0$
$a = 3 , b = - 4 , c = - 1$

$D = {b}^{2} - 4 a c$
$\Rightarrow D = {\left(- 4\right)}^{2} - 4 \left(3\right) \left(- 1\right)$
$\Rightarrow D = 16 + 12$
$\Rightarrow D = 28$

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