12. $sqrt{x + 4}+x<6$

Explanation:

Step1: Isolate the square - root term

Subtract (x) from both sides of the inequality (\sqrt{x + 4}+x<6) to get (\sqrt{x + 4}<6 - x).

Step2: Consider the domain of the square - root

For (\sqrt{x + 4}) to be a real number, we must have (x+4\geq0), which implies (x\geq - 4). Also, since (\sqrt{x + 4}<6 - x), we need (6 - x>0) (because the square - root is non - negative), so (x<6).

Step3: Square both sides of the inequality

Squaring both sides of (\sqrt{x + 4}<6 - x) (note that since (6 - x>0), squaring is valid), we get (x + 4<(6 - x)^2). Expand ((6 - x)^2) to (36-12x+x^{2}). So (x + 4<36-12x+x^{2}).

Step4: Rearrange to form a quadratic inequality

Rearrange (x + 4<36-12x+x^{2}) to (x^{2}-13x + 32>0). The roots of the quadratic equation (x^{2}-13x + 32 = 0) are given by (x=\frac{13\pm\sqrt{13^{2}-4\times32}}{2}=\frac{13\pm\sqrt{169 - 128}}{2}=\frac{13\pm\sqrt{41}}{2}).

Step5: Find the solution of the quadratic inequality

The quadratic function (y=x^{2}-13x + 32) is a parabola opening upwards. The inequality (x^{2}-13x + 32>0) is satisfied when (x<\frac{13-\sqrt{41}}{2}) or (x>\frac{13 + \sqrt{41}}{2}).

Step6: Combine with the domain

Combining with the domain ( - 4\leq x<6), we find that the solution of the original inequality is (-4\leq x<\frac{13-\sqrt{41}}{2}).

Answer:

(-4\leq x<\frac{13 - \sqrt{41}}{2})