The roots of $x^2+(2k+1)x+(k-3)=0$ are real and opposite in sign exactly when:

Roots of opposite sign means their product is negative, and the product is the constant term $k-3$, so $k<3$. Reality needs no separate check: the discriminant $(2k+1{)}^2-4(k-3)=4k^2+13$ is positive for every $k.(k>3$ makes both roots share a sign; the two options built on −1/2 come from the sum $-(2k+1)$, which says where the roots sit, not whether they straddle zero.) At $k=0$ the equation $x^2+x-3=0$ has roots ≈ 1.30 and ≈ −2.30.