\(2x^2+kx+5=0 \\ \\
a=2 \\ b=k \\ c=5 \\ \Delta=b^2-4ac=k^2-4 \times 2 \times 5=k^2-40\)
When the discriminant Δ is greater than 0, the equation has 2 real solutions:
\(k^2-40 > 0 \\
k^2>40 \\
k>\sqrt{40} \ \lor \ k<-\sqrt{40} \\
k>2\sqrt{10} \ \lor \ k<-2\sqrt{10} \\
\boxed{k \in (-\infty,-2\sqrt{10}) \cup (2\sqrt{10}, +\infty)}\)
When the discriminant Δ is less than 0, the equation has 2 complex and no real solutions:
\(k^2-40<0 \\
k^2<40 \\
k<\sqrt{40} \ \land \ k>-\sqrt{40} \\
k<2\sqrt{10} \ \land \ k>-2\sqrt{10} \\
\boxed{k \in (-2\sqrt{10},2\sqrt{10})}\)
When the discriminant Δ is equal to 0, the equation has exactly 1 real solution:
\(k^2-40=0 \\
k^2=40 \\
k=\sqrt{40} \ \lor \ k=-\sqrt{40} \\
k=2\sqrt{10} \ \lor \ k=-2\sqrt{10} \\
\boxed{k \in \{ -2\sqrt{10}, 2\sqrt{10} \} }\)
