Solve (2x+5)(5-2x)+(2x-3)^2-2>0

$$( 2 x + 5 ) ( 5 - 2 x ) + ( 2 x - 3 ) ^ { 2 } - 2 > 0$$

Solve for x

$$x<\frac{8}{3}$$

Solution Steps

Consider \left(2x+5\right)\left(5-2x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 5.

$$25-\left(2x\right)^{2}+\left(2x-3\right)^{2}-2>0$$

Expand \left(2x\right)^{2}.

$$25-2^{2}x^{2}+\left(2x-3\right)^{2}-2>0$$

Calculate 2 to the power of 2 and get 4.

$$25-4x^{2}+\left(2x-3\right)^{2}-2>0$$

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

$$25-4x^{2}+4x^{2}-12x+9-2>0$$

Combine -4x^{2} and 4x^{2} to get 0.

$$25-12x+9-2>0$$

Add 25 and 9 to get 34.

$$34-12x-2>0$$

Subtract 2 from 34 to get 32.

$$32-12x>0$$

Subtract 32 from both sides. Anything subtracted from zero gives its negation.

$$-12x>-32$$

Divide both sides by -12. Since -12 is negative, the inequality direction is changed.

$$x<\frac{-32}{-12}$$

Reduce the fraction \frac{-32}{-12} to lowest terms by extracting and canceling out -4.

$$x<\frac{8}{3}$$