Solve -4x^2+16x+272>0

$$-4 { x }^{ 2 } +16x+272 > 0$$

Solve for x

$$x\in \left(2-6\sqrt{2},6\sqrt{2}+2\right)$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in -4x^{2}+16x+272 positive. Since -1 is negative, the inequality direction is changed.

$$4x^{2}-16x-272<0$$

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.

$$4x^{2}-16x-272=0$$

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 4 for a, -16 for b, and -272 for c in the quadratic formula.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 4\left(-272\right)}}{2\times 4}$$

Do the calculations.

$$x=\frac{16±48\sqrt{2}}{8}$$

Solve the equation x=\frac{16±48\sqrt{2}}{8} when ± is plus and when ± is minus.

$$x=6\sqrt{2}+2 x=2-6\sqrt{2}$$

Rewrite the inequality by using the obtained solutions.

$$4\left(x-\left(6\sqrt{2}+2\right)\right)\left(x-\left(2-6\sqrt{2}\right)\right)<0$$

For the product to be negative, x-\left(6\sqrt{2}+2\right) and x-\left(2-6\sqrt{2}\right) have to be of the opposite signs. Consider the case when x-\left(6\sqrt{2}+2\right) is positive and x-\left(2-6\sqrt{2}\right) is negative.

$$x-\left(6\sqrt{2}+2\right)>0 x-\left(2-6\sqrt{2}\right)<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\left(2-6\sqrt{2}\right) is positive and x-\left(6\sqrt{2}+2\right) is negative.

$$x-\left(2-6\sqrt{2}\right)>0 x-\left(6\sqrt{2}+2\right)<0$$

The solution satisfying both inequalities is x\in \left(2-6\sqrt{2},6\sqrt{2}+2\right).

$$x\in \left(2-6\sqrt{2},6\sqrt{2}+2\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(2-6\sqrt{2},6\sqrt{2}+2\right)$$