Solve $x^{2} + 6 x + 13 > 0$ .

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Solution

Given inequation is $x^{2} + 6 x + 13 > 0$ .

Here, factorization is not possible.

Rewriting the given inequation, we get,
$(x^{2} + 6 x + 9) + 4 > 0$

$(x + 3)^{2} + 4 > 0$

We know that $(x + 3)^{2} \geq 0$ for every $x \in R$ .
$(x + 3)^{2} + 4 \geq 4 > 0$ for every $x \in R$ .

Therefore, the required solution is the set of all real numbers, i.e., $(- \infty, \infty)$ .

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