Solve x^2+8x+16>0

$${ x }^{ 2 } +8x+16 > 0$$

Solve for x

$$x\neq -4$$

Steps Using the Quadratic Formula

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.

$$x^{2}+8x+16=0$$

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, 8 for b, and 16 for c in the quadratic formula.

$$x=\frac{-8±\sqrt{8^{2}-4\times 1\times 16}}{2}$$

Do the calculations.

$$x=\frac{-8±0}{2}$$

Solutions are the same.

$$x=-4$$

Rewrite the inequality by using the obtained solutions.

$$\left(x+4\right)^{2}>0$$

Inequality holds for x\neq -4.

$$x\neq -4$$