Solve (2x-1)^2+2(2x-1)-3>0

$$( 2 x - 1 ) ^ { 2 } + 2 ( 2 x - 1 ) - 3 > 0$$

Solve for x

$$x\in \left(-\infty,-1\right)\cup \left(1,\infty\right)$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-1\right)^{2}.

$$4x^{2}-4x+1+2\left(2x-1\right)-3>0$$

Use the distributive property to multiply 2 by 2x-1.

$$4x^{2}-4x+1+4x-2-3>0$$

Combine -4x and 4x to get 0.

$$4x^{2}+1-2-3>0$$

Subtract 2 from 1 to get -1.

$$4x^{2}-1-3>0$$

Subtract 3 from -1 to get -4.

$$4x^{2}-4>0$$

Add 1 to both sides.

$$x^{2}>1$$

Calculate the square root of 1 and get 1. Rewrite 1 as 1^{2}.

$$x^{2}>1^{2}$$

Inequality holds for |x|>1.

$$|x|>1$$

Rewrite |x|>1 as x<-1\text{; }x>1.

$$x<-1\text{; }x>1$$