Solve for d 2n^2d=fd+m

$2 n^{2} d = f d + m$

Step 1

Subtract $f d$ from both sides of the equation.

$2 n^{2} d - f d = m$

Step 2

Factor $d$ out of $2 n^{2} d - f d$ .

Factor $d$ out of $2 n^{2} d$ .

$d (2 n^{2}) - f d = m$

Factor $d$ out of $- f d$ .

$d (2 n^{2}) + d (- f) = m$

Factor $d$ out of $d (2 n^{2}) + d (- f)$ .

$d (2 n^{2} - f) = m$

Step 3

Divide each term in $d (2 n^{2} - f) = m$ by $2 n^{2} - f$ and simplify.

Divide each term in $d (2 n^{2} - f) = m$ by $2 n^{2} - f$ .

$\frac{d (2 n^{2} - f)}{2 n^{2} - f} = \frac{m}{2 n^{2} - f}$

Simplify the left side.

Cancel the common factor of $2 n^{2} - f$ .

Cancel the common factor.

$\frac{d (2 n^{2} - f)}{2 n^{2} - f} = \frac{m}{2 n^{2} - f}$

Divide $d$ by $1$ .

$d = \frac{m}{2 n^{2} - f}$