Solve left(2x-1right)^2-4(x^2+3)>0

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

Solve for x

$$x<-\frac{11}{4}$$

Solution Steps

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

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

Use the distributive property to multiply -4 by x^{2}+3.

$$4x^{2}-4x+1-4x^{2}-12>0$$

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

$$-4x+1-12>0$$

Subtract 12 from 1 to get -11.

$$-4x-11>0$$

Add 11 to both sides. Anything plus zero gives itself.

$$-4x>11$$

Divide both sides by -4. Since -4 is negative, the inequality direction is changed.

$$x<-\frac{11}{4}$$