Solve x^2-4kx+k+3=0

$$x ^ { 2 } - 4 k x + k + 3 = 0$$

Solve for k

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

$$x\neq \frac{1}{4}$$

Steps for Solving Linear Equation

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

$$-4kx+k+3=-x^{2}$$

Subtract 3 from both sides.

$$-4kx+k=-x^{2}-3$$

Combine all terms containing k.

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

The equation is in standard form.

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

Divide both sides by -4x+1.

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

Dividing by -4x+1 undoes the multiplication by -4x+1.

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

Divide -x^{2}-3 by -4x+1.

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

Solve for x

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

$$x=-\sqrt{\left(k-1\right)\left(4k+3\right)}+2k\text{, }k\leq -\frac{3}{4}\text{ or }k\geq 1$$