How do you solve #2(x+3)^2 + 5 > 0#?
This is the vertex form of the parabola graph.
Vertex (-3, 5). The parabola opens upward (a > 0). The parabola never intersects the x-axis. The inequality is always true.
Another way to prove it is to convert f(x) into standard form
f(x) = 2x^2 + 12x + 23 > 0
D = b^2 - 4ac = 144 - 184 = -40 < 0.
Then, f(x) has the same sign with a = 2 > 0. The inequality is always true, for whatever value of x.