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

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

Solve for x

$$x>-3$$

Solution Steps

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

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

Add 1 and 1 to get 2.

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

Subtract 4x^{2} from both sides.

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

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

$$4x+2>3x-1$$

Subtract 3x from both sides.

$$4x+2-3x>-1$$

Combine 4x and -3x to get x.

$$x+2>-1$$

Subtract 2 from both sides.

$$x>-1-2$$

Subtract 2 from -1 to get -3.

$$x>-3$$