Solve the Quadratic Inequality: Where f(x)=-2x²+4x-6 Is Greater Than 0

The following function is graphed below:

$f(x)=-2x^2+4x-6$

For which values of x is

$f(x)>0$ true?

The given function is $f(x) = -2x^2 + 4x - 6$.

First, we find the roots of the equation $f(x) = 0$, which are the critical points where the function can change its sign.

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = -2$, $b = 4$, and $c = -6$:

Calculate the discriminant $\Delta = b^2 - 4ac = 4^2 - 4(-2)(-6) = 16 - 48 = -32$.

The discriminant is negative, indicating there are no real roots.

Because the parabola opens downwards and there are no x-intercepts (real roots), the function does not cross the x-axis.

Hence, for all real numbers, $f(x) \leq 0$.

Therefore, there are no values of $x$ for which $f(x) > 0$.

Thus, the solution is: No answer.