Solve the inequality -x^2 + 6x - 10 > 0.

Step-by-step solution

1. 1

   Flip to a positive leading coefficient. Multiply by $-1$ and reverse the sign:

   $x^2-6x+10<0$

2. 2

   Test for real roots with the discriminant. With $a=1$, $b=-6$, $c=10$:

   $\Delta=(-6)^2-4(1)(10)=36-40=-4<0$

   A negative discriminant means the parabola never crosses the $x$-axis, so the expression never changes sign.

3. 3

   Determine that single sign. Since $a=1>0$ the parabola opens upward and lies entirely above the axis. Completing the square makes it explicit:

   $x^2-6x+10=(x-3)^2+1\ge 1>0$

   The minimum value is $1$, attained at $x=3$.

4. 4

   Conclude. An expression that is always at least $1$ can never be $<0$, so the solution set is empty:

   $x\in\varnothing$

5. 5

   Check the original form directly. $-x^2+6x-10=-\left[(x-3)^2+1\right]\le-1<0$ for every real $x$, so it is never positive $\checkmark$. Spot-check the vertex $x=3$: $-9+18-10=-1$, the largest value the left-hand side ever takes.