Solve -4x^2+(1-k)x-1=0

$$- 4 x ^ { 2 } + ( 1 - k ) x - 1 = 0$$

Solve for k

$$k=-4x+1-\frac{1}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

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

$$-4x^{2}+x-kx-1=0$$

Add 4x^{2} to both sides. Anything plus zero gives itself.

$$x-kx-1=4x^{2}$$

Subtract x from both sides.

$$-kx-1=4x^{2}-x$$

Add 1 to both sides.

$$-kx=4x^{2}-x+1$$

The equation is in standard form.

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

Divide both sides by -x.

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

Dividing by -x undoes the multiplication by -x.

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

Divide 4x^{2}-x+1 by -x.

$$k=-4x+1-\frac{1}{x}$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{\left(k-5\right)\left(k+3\right)}-k+1}{8}$$

$$x=\frac{\sqrt{\left(k-5\right)\left(k+3\right)}-k+1}{8}$$

Solve for x

$$x=\frac{-\sqrt{\left(k-5\right)\left(k+3\right)}-k+1}{8}$$

$$x=\frac{\sqrt{\left(k-5\right)\left(k+3\right)}-k+1}{8}\text{, }k\leq -3\text{ or }k\geq 5$$