Easy points are not easy to get

$x=\dfrac { -3\pm \sqrt { 9+4k } }{ 2k }$ For this to be real, the expression inside roots must be $\geq 0$ . Therefore,

$9+4k \geq 0$

$k + \dfrac{9}{4} \geq 0$

$\therefore k \in \left[-\dfrac{9}{4}, \infty\right]$ $\implies \text{Least Value = }-\dfrac{9}{4} = \boxed{\large{-2.25}}$

quadratic equation have real solution if D=b^2 - 4ac >=0 (not negatif)

kx^2+3x-1 = 0 => a=k, b=3, and c=-1

D=b^2-4ac=3^2 - 4 k (-1)=9+4k

D>=0

9+4k>=0

k>=(-9)/4

k>=-2,25

so, small value for k is -2,25