Solve -x^2+6x-8<0

$$- x ^ { 2 } + 6 x - 8 < 0$$

Solve for x

$$x\in \left(-\infty,2\right)\cup \left(4,\infty\right)$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in -x^{2}+6x-8 positive. Since -1 is negative, the inequality direction is changed.

$$x^{2}-6x+8>0$$

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$x^{2}-6x+8=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, -6 for b, and 8 for c in the quadratic formula.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 1\times 8}}{2}$$

Do the calculations.

$$x=\frac{6±2}{2}$$

Solve the equation x=\frac{6±2}{2} when ± is plus and when ± is minus.

$$x=4 x=2$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-4\right)\left(x-2\right)>0$$

For the product to be positive, x-4 and x-2 have to be both negative or both positive. Consider the case when x-4 and x-2 are both negative.

$$x-4<0 x-2<0$$

The solution satisfying both inequalities is x<2.

$$x<2$$

Consider the case when x-4 and x-2 are both positive.

$$x-2>0 x-4>0$$

The solution satisfying both inequalities is x>4.

$$x>4$$

The final solution is the union of the obtained solutions.

$$x<2\text{; }x>4$$