Solve Negative Inequality in Quadratic: y=-x²+6x-8

To solve the problem of finding where the function $y = -x^2 + 6x - 8$ is less than zero, we follow these steps:

• Step 1: Set the function equal to zero and use the quadratic formula to find the roots.
• Step 2: Analyze the sign of the function between and beyond the roots to identify the intervals where the function is negative.

Let's work through each step:
Step 1: The function $y = -x^2 + 6x - 8$ can be set to 0:
$-x^2 + 6x - 8 = 0$

Using the quadratic formula where $a = -1$, $b = 6$, and $c = -8$:
Discriminant $D = b^2 - 4ac = 6^2 - 4(-1)(-8) = 36 - 32 = 4$

The roots are:
$x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-6 \pm \sqrt{4}}{-2} = \frac{-6 \pm 2}{-2}$

The solutions are:
$x = \frac{-6 + 2}{-2} = 2$ and $x = \frac{-6 - 2}{-2} = 4$

Step 2: Determine where the function is negative. Since the parabola opens downwards, it will be negative outside of the roots.
Therefore, the function is negative for:
$x < 2$ and $x > 4$

Therefore, the solution to the problem is:

$x < 2$ or $x > 4$