Solve for b N=-m^2+(3d)/b

$N = - m^{2} + \frac{3 d}{b}$

Step 1

Rewrite the equation as $- m^{2} + \frac{3 d}{b} = N$ .

$- m^{2} + \frac{3 d}{b} = N$

Step 2

Add $m^{2}$ to both sides of the equation.

$\frac{3 d}{b} = N + m^{2}$

Step 3

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$b, 1, 1$

The LCM of one and any expression is the expression.

$b$

Step 4

Multiply each term in $\frac{3 d}{b} = N + m^{2}$ by $b$ to eliminate the fractions.

Multiply each term in $\frac{3 d}{b} = N + m^{2}$ by $b$ .

$\frac{3 d}{b} b = N b + m^{2} b$

Simplify the left side.

Cancel the common factor of $b$ .

Cancel the common factor.

$\frac{3 d}{b} b = N b + m^{2} b$

Rewrite the expression.

$3 d = N b + m^{2} b$

Step 5

Solve the equation.

Rewrite the equation as $N b + m^{2} b = 3 d$ .

$N b + m^{2} b = 3 d$

Factor $b$ out of $N b + m^{2} b$ .

Factor $b$ out of $N b$ .

$b N + m^{2} b = 3 d$

Factor $b$ out of $m^{2} b$ .

$b N + b m^{2} = 3 d$

Factor $b$ out of $b N + b m^{2}$ .

$b (N + m^{2}) = 3 d$

Divide each term in $b (N + m^{2}) = 3 d$ by $N + m^{2}$ and simplify.

Divide each term in $b (N + m^{2}) = 3 d$ by $N + m^{2}$ .

$\frac{b (N + m^{2})}{N + m^{2}} = \frac{3 d}{N + m^{2}}$

Simplify the left side.

Cancel the common factor of $N + m^{2}$ .

Cancel the common factor.

$\frac{b (N + m^{2})}{N + m^{2}} = \frac{3 d}{N + m^{2}}$

Divide $b$ by $1$ .

$b = \frac{3 d}{N + m^{2}}$