quadratic

If $x^2 + 2ax + 10 - 3a > 0$ for all real numbers x, then this means the graph of $f(x) = x^2 + 2ax + 10 - 3a$ never touches the x-axis. Because if it touches the x-axis, at that point, the function would be zero. However, the questions says that f(x) > 0, so it is never zero, implying it never touches the x-axis.

For quadratic functions that doesn't touch the x-axis, we always have $\Delta < 0$, where $\Delta$ is the discriminant.

We calculate $\Delta$.

$\Delta = (2a)^2 - 4(1)(10 - 3a) = 4a^2 + 12a - 40 = 4(a^2 + 3a - 10)$

If we solve the inequality $\Delta < 0$, we get $-5 < a < 2$ as our answer. (how?)

x=−a+√[(a−2)(a+5)]

x=−a−√[(a−2)(a+5)]

so x can be anything when x^(2) + 2ax- 3a >-10 is true where we substitute x with the equation above to get x=(-5)<=>(2)