Solve (3x-2)^2-8x^2<(x+4)(x-7)

$$( 3 x - 2 ) ^ { 2 } - 8 x ^ { 2 } < ( x + 4 ) ( x - 7 )$$

Solve for x

$$x>\frac{32}{9}$$

Solution Steps

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

$$9x^{2}-12x+4-8x^{2}<\left(x+4\right)\left(x-7\right)$$

Combine 9x^{2} and -8x^{2} to get x^{2}.

$$x^{2}-12x+4<\left(x+4\right)\left(x-7\right)$$

Use the distributive property to multiply x+4 by x-7 and combine like terms.

$$x^{2}-12x+4<x^{2}-3x-28$$

Subtract x^{2} from both sides.

$$x^{2}-12x+4-x^{2}<-3x-28$$

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

$$-12x+4<-3x-28$$

Add 3x to both sides.

$$-12x+4+3x<-28$$

Combine -12x and 3x to get -9x.

$$-9x+4<-28$$

Subtract 4 from both sides.

$$-9x<-28-4$$

Subtract 4 from -28 to get -32.

$$-9x<-32$$

Divide both sides by -9. Since -9 is negative, the inequality direction is changed.

$$x>\frac{-32}{-9}$$

Fraction \frac{-32}{-9} can be simplified to \frac{32}{9} by removing the negative sign from both the numerator and the denominator.

$$x>\frac{32}{9}$$