Solve x^2+4x+8>0

$$x ^ { 2 } + 4 x + 8 > 0$$

Solve for x

$$x\in \mathrm{R}$$

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.

$$x^{2}+4x+8=0$$

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

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

Do the calculations.

$$x=\frac{-4±\sqrt{-16}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Expression x^{2}+4x+8 has the same sign for any x. To determine the sign, calculate the value of the expression for x=0.

$$0^{2}+4\times 0+8=8$$

The value of the expression x^{2}+4x+8 is always positive. Inequality holds for x\in \mathrm{R}.

$$x\in \mathrm{R}$$