Solve f(2x-3)*(2x+1)-4x^2=1

$$f ( 2 x - 3 ) \cdot ( 2 x + 1 ) - 4 x ^ { 2 } = 1$$

Solve for f

$$f=\frac{4x^{2}+1}{4x^{2}-4x-3}$$

$$x\neq -\frac{1}{2}\text{ and }x\neq \frac{3}{2}$$

Steps for Solving Linear Equation

Use the distributive property to multiply f by 2x-3.

$$\left(2fx-3f\right)\left(2x+1\right)-4x^{2}=1$$

Use the distributive property to multiply 2fx-3f by 2x+1 and combine like terms.

$$4fx^{2}-4fx-3f-4x^{2}=1$$

Add 4x^{2} to both sides.

$$4fx^{2}-4fx-3f=1+4x^{2}$$

Combine all terms containing f.

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

The equation is in standard form.

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

Divide both sides by 4x^{2}-4x-3.

$$\frac{\left(4x^{2}-4x-3\right)f}{4x^{2}-4x-3}=\frac{4x^{2}+1}{4x^{2}-4x-3}$$

Dividing by 4x^{2}-4x-3 undoes the multiplication by 4x^{2}-4x-3.

$$f=\frac{4x^{2}+1}{4x^{2}-4x-3}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{\sqrt{4f^{2}-2f-1}+f}{2\left(f-1\right)}\text{; }x=\frac{-\sqrt{4f^{2}-2f-1}+f}{2\left(f-1\right)}\text{, }&f\neq 1\\x=-1\text{, }&f=1\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{\sqrt{4f^{2}-2f-1}+f}{2\left(f-1\right)}\text{; }x=\frac{-\sqrt{4f^{2}-2f-1}+f}{2\left(f-1\right)}\text{, }&f\leq \frac{1-\sqrt{5}}{4}\text{ or }\left(f\neq 1\text{ and }f\geq \frac{\sqrt{5}+1}{4}\right)\\x=-1\text{, }&f=1\end{matrix}\right.$$