Solve n(m+1)(2n+1)/6=91

$$\frac { n ( m + 1 ) ( 2 n + 1 ) } { 6 } = 91$$

Solve for m

$$m=-\frac{2n^{2}+n-546}{n\left(2n+1\right)}$$

$$n\neq -\frac{1}{2}\text{ and }n\neq 0$$

Steps for Solving Linear Equation

Multiply both sides by 6.

$$n\left(m+1\right)\left(2n+1\right)=91\times 6$$

Use the distributive property to multiply n by m+1.

$$\left(nm+n\right)\left(2n+1\right)=91\times 6$$

Use the distributive property to multiply nm+n by 2n+1.

$$2mn^{2}+nm+2n^{2}+n=91\times 6$$

Multiply 91 and 6 to get 546.

$$2mn^{2}+nm+2n^{2}+n=546$$

Subtract 2n^{2} from both sides.

$$2mn^{2}+nm+n=546-2n^{2}$$

Subtract n from both sides.

$$2mn^{2}+nm=546-2n^{2}-n$$

Combine all terms containing m.

$$\left(2n^{2}+n\right)m=546-2n^{2}-n$$

The equation is in standard form.

$$\left(2n^{2}+n\right)m=546-n-2n^{2}$$

Divide both sides by 2n^{2}+n.

$$\frac{\left(2n^{2}+n\right)m}{2n^{2}+n}=\frac{546-n-2n^{2}}{2n^{2}+n}$$

Dividing by 2n^{2}+n undoes the multiplication by 2n^{2}+n.

$$m=\frac{546-n-2n^{2}}{2n^{2}+n}$$

Divide 546-2n^{2}-n by 2n^{2}+n.

$$m=\frac{546-n-2n^{2}}{n\left(2n+1\right)}$$

Solve for n

$$n=\frac{\sqrt{\left(m+1\right)\left(m+4369\right)}-m-1}{4\left(m+1\right)}$$

$$n=-\frac{\sqrt{\left(m+1\right)\left(m+4369\right)}+m+1}{4\left(m+1\right)}\text{, }m>-1\text{ or }m\leq -4369$$