Solve (x-3)^2+3(3x+2)>(x+6)(x+3)+12

$$( x - 3 ) ^ { 2 } + 3 ( 3 x + 2 ) > ( x + 6 ) ( x + 3 ) + 12$$

Solve for x

$$x<-\frac{5}{2}$$

Solution Steps

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

$$x^{2}-6x+9+3\left(3x+2\right)>\left(x+6\right)\left(x+3\right)+12$$

Use the distributive property to multiply 3 by 3x+2.

$$x^{2}-6x+9+9x+6>\left(x+6\right)\left(x+3\right)+12$$

Combine -6x and 9x to get 3x.

$$x^{2}+3x+9+6>\left(x+6\right)\left(x+3\right)+12$$

Add 9 and 6 to get 15.

$$x^{2}+3x+15>\left(x+6\right)\left(x+3\right)+12$$

Use the distributive property to multiply x+6 by x+3 and combine like terms.

$$x^{2}+3x+15>x^{2}+9x+18+12$$

Add 18 and 12 to get 30.

$$x^{2}+3x+15>x^{2}+9x+30$$

Subtract x^{2} from both sides.

$$x^{2}+3x+15-x^{2}>9x+30$$

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

$$3x+15>9x+30$$

Subtract 9x from both sides.

$$3x+15-9x>30$$

Combine 3x and -9x to get -6x.

$$-6x+15>30$$

Subtract 15 from both sides.

$$-6x>30-15$$

Subtract 15 from 30 to get 15.

$$-6x>15$$

Divide both sides by -6. Since -6 is negative, the inequality direction is changed.

$$x<\frac{15}{-6}$$

Reduce the fraction \frac{15}{-6} to lowest terms by extracting and canceling out 3.

$$x<-\frac{5}{2}$$