Solve x^2+2x+6leq6+left(3+xright)^2

$${ x }^{ 2 } +2x+6 \leq 6+ { \left(3+x \right) }^{ 2 }$$

Solve for x

$$x\geq -\frac{9}{4}$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3+x\right)^{2}.

$$x^{2}+2x+6\leq 6+9+6x+x^{2}$$

Add 6 and 9 to get 15.

$$x^{2}+2x+6\leq 15+6x+x^{2}$$

Subtract 6x from both sides.

$$x^{2}+2x+6-6x\leq 15+x^{2}$$

Combine 2x and -6x to get -4x.

$$x^{2}-4x+6\leq 15+x^{2}$$

Subtract x^{2} from both sides.

$$x^{2}-4x+6-x^{2}\leq 15$$

Combine x^{2} and -x^{2} to get 0.

$$-4x+6\leq 15$$

Subtract 6 from both sides.

$$-4x\leq 15-6$$

Subtract 6 from 15 to get 9.

$$-4x\leq 9$$

Divide both sides by -4. Since -4 is negative, the inequality direction is changed.

$$x\geq -\frac{9}{4}$$