Solve 2(p^2+8p+19)>0

$$2 ( p ^ { 2 } + 8 p + 19 ) > 0$$

Solve for p

$$p\in \mathrm{R}$$

Steps Using the Quadratic Formula

Divide both sides by 2. Since 2 is positive, the inequality direction remains the same. Zero divided by any non-zero number gives zero.

$$p^{2}+8p+19>0$$

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.

$$p^{2}+8p+19=0$$

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 19 for c in the quadratic formula.

$$p=\frac{-8±\sqrt{8^{2}-4\times 1\times 19}}{2}$$

Do the calculations.

$$p=\frac{-8±\sqrt{-12}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Expression p^{2}+8p+19 has the same sign for any p. To determine the sign, calculate the value of the expression for p=0.

$$0^{2}+8\times 0+19=19$$

The value of the expression p^{2}+8p+19 is always positive. Inequality holds for p\in \mathrm{R}.

$$p\in \mathrm{R}$$