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

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

Solve for λ

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

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

Steps for Solving Linear Equation

Multiply both sides by 6.

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

Multiply 91 and 6 to get 546.

$$546=n\left(\lambda +1\right)\left(2n+1\right)$$

Use the distributive property to multiply n by \lambda +1.

$$546=\left(n\lambda +n\right)\left(2n+1\right)$$

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

$$546=2\lambda n^{2}+n\lambda +2n^{2}+n$$

Swap sides so that all variable terms are on the left hand side.

$$2\lambda n^{2}+n\lambda +2n^{2}+n=546$$

Subtract 2n^{2} from both sides.

$$2\lambda n^{2}+n\lambda +n=546-2n^{2}$$

Subtract n from both sides.

$$2\lambda n^{2}+n\lambda =546-2n^{2}-n$$

Combine all terms containing \lambda .

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

The equation is in standard form.

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

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

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

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

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

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

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

Solve for n

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

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