Solve 16x^2+(2k+1)x+k-1=0

$$16 x ^ { 2 } + ( 2 k + 1 ) x + k - 1 = 0$$

Solve for k

$$k=-\frac{16x^{2}+x-1}{2x+1}$$

$$x\neq -\frac{1}{2}$$

Steps for Solving Linear Equation

Use the distributive property to multiply 2k+1 by x.

$$16x^{2}+2kx+x+k-1=0$$

Subtract 16x^{2} from both sides. Anything subtracted from zero gives its negation.

$$2kx+x+k-1=-16x^{2}$$

Subtract x from both sides.

$$2kx+k-1=-16x^{2}-x$$

Add 1 to both sides.

$$2kx+k=-16x^{2}-x+1$$

Combine all terms containing k.

$$\left(2x+1\right)k=-16x^{2}-x+1$$

The equation is in standard form.

$$\left(2x+1\right)k=1-x-16x^{2}$$

Divide both sides by 2x+1.

$$\frac{\left(2x+1\right)k}{2x+1}=\frac{1-x-16x^{2}}{2x+1}$$

Dividing by 2x+1 undoes the multiplication by 2x+1.

$$k=\frac{1-x-16x^{2}}{2x+1}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{4k^{2}-60k+65}}{32}-\frac{k}{16}-\frac{1}{32}$$

$$x=-\frac{\sqrt{4k^{2}-60k+65}}{32}-\frac{k}{16}-\frac{1}{32}$$

Solve for x

$$x=\frac{\sqrt{4k^{2}-60k+65}}{32}-\frac{k}{16}-\frac{1}{32}$$

$$x=-\frac{\sqrt{4k^{2}-60k+65}}{32}-\frac{k}{16}-\frac{1}{32}\text{, }k\geq 2\sqrt{10}+\frac{15}{2}\text{ or }k\leq \frac{15}{2}-2\sqrt{10}$$