What is the answer to solvefor d,2n^2d=fd+m ?

The answer to solvefor d,2n^2d=fd+m is d= m/(2n^2-f);\quad n\ne sqrt(f/2),n\ne-sqrt(f/2)

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solvefor d,2n^2d=fd+m

solve for

d, 2 n^{2} d = fd + m

d = \frac{m}{2 n ^{2} - f}; n \neq = \frac{f}{2}, n \neq = - \frac{f}{2}

Solution steps

2 n^{2} d = fd + m

Move

fd

to the left side

2 n^{2} d - fd = m

Factor

2 n^{2} d - fd : d (2 n^{2} - f)

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

Divide both sides by

2 n^{2} - f; n \neq = \frac{f}{2}, n \neq = - \frac{f}{2}

d = \frac{m}{2 n ^{2} - f}; n \neq = \frac{f}{2}, n \neq = - \frac{f}{2}

x, 3 z + 8 = 12 + 3 x - z

2 (n + 6) - 2 = 4

- 9 x + 2 = - 43

- 2 = \frac{b}{- 10}

- 5 (- a - 1) = 3 a - 7

What is the answer to solvefor d,2n^2d=fd+m ?
The answer to solvefor d,2n^2d=fd+m is d= m/(2n^2-f);\quad n\ne sqrt(f/2),n\ne-sqrt(f/2)