Solve 4mk+2k^2-k=65

$$4 m k + 2 k ^ { 2 } - k = 65$$

Solve for m

$$m=-\frac{k}{2}+\frac{1}{4}+\frac{65}{4k}$$

$$k\neq 0$$

Steps for Solving Linear Equation

Subtract 2k^{2} from both sides.

$$4mk-k=65-2k^{2}$$

Add k to both sides.

$$4mk=65-2k^{2}+k$$

The equation is in standard form.

$$4km=65+k-2k^{2}$$

Divide both sides by 4k.

$$\frac{4km}{4k}=\frac{65+k-2k^{2}}{4k}$$

Dividing by 4k undoes the multiplication by 4k.

$$m=\frac{65+k-2k^{2}}{4k}$$

Divide 65-2k^{2}+k by 4k.

$$m=-\frac{k}{2}+\frac{1}{4}+\frac{65}{4k}$$

Solve for k

$$k=\frac{\sqrt{16m^{2}-8m+521}}{4}-m+\frac{1}{4}$$

$$k=-\frac{\sqrt{16m^{2}-8m+521}}{4}-m+\frac{1}{4}$$