Solve left(x+3right)^2+6leqx^2-3x

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

Solve for x

$$x\leq -\frac{5}{3}$$

Solution Steps

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

$$x^{2}+6x+9+6\leq x^{2}-3x$$

Add 9 and 6 to get 15.

$$x^{2}+6x+15\leq x^{2}-3x$$

Subtract x^{2} from both sides.

$$x^{2}+6x+15-x^{2}\leq -3x$$

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

$$6x+15\leq -3x$$

Add 3x to both sides.

$$6x+15+3x\leq 0$$

Combine 6x and 3x to get 9x.

$$9x+15\leq 0$$

Subtract 15 from both sides. Anything subtracted from zero gives its negation.

$$9x\leq -15$$

Divide both sides by 9. Since 9 is positive, the inequality direction remains the same.

$$x\leq \frac{-15}{9}$$

Reduce the fraction \frac{-15}{9} to lowest terms by extracting and canceling out 3.

$$x\leq -\frac{5}{3}$$