Solve -x^2+6x+40>0

$$- x ^ { 2 } + 6 x + 40 > 0$$

Solve for x

$$x\in \left(-4,10\right)$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in -x^{2}+6x+40 positive. Since -1 is negative, the inequality direction is changed.

$$x^{2}-6x-40<0$$

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.

$$x^{2}-6x-40=0$$

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 -40 for c in the quadratic formula.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 1\left(-40\right)}}{2}$$

Do the calculations.

$$x=\frac{6±14}{2}$$

Solve the equation x=\frac{6±14}{2} when ± is plus and when ± is minus.

$$x=10 x=-4$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-10\right)\left(x+4\right)<0$$

For the product to be negative, x-10 and x+4 have to be of the opposite signs. Consider the case when x-10 is positive and x+4 is negative.

$$x-10>0 x+4<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x+4 is positive and x-10 is negative.

$$x+4>0 x-10<0$$

The solution satisfying both inequalities is x\in \left(-4,10\right).

$$x\in \left(-4,10\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(-4,10\right)$$