Solve 9n^2+10=91

$$9 n ^ { 2 } + 10 = 91$$

Solve for n

$$n=3$$

$$n=-3$$

Steps Using Difference of Squares

Subtract 91 from both sides.

$$9n^{2}+10-91=0$$

Subtract 91 from 10 to get -81.

$$9n^{2}-81=0$$

Divide both sides by 9.

$$n^{2}-9=0$$

Consider n^{2}-9. Rewrite n^{2}-9 as n^{2}-3^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(n-3\right)\left(n+3\right)=0$$

To find equation solutions, solve n-3=0 and n+3=0.

$$n=3 n=-3$$

Steps by Finding Square Root

Subtract 10 from both sides.

$$9n^{2}=91-10$$

Subtract 10 from 91 to get 81.

$$9n^{2}=81$$

Divide both sides by 9.

$$n^{2}=\frac{81}{9}$$

Divide 81 by 9 to get 9.

$$n^{2}=9$$

Take the square root of both sides of the equation.

$$n=3 n=-3$$

Steps Using the Quadratic Formula

Subtract 91 from both sides.

$$9n^{2}+10-91=0$$

Subtract 91 from 10 to get -81.

$$9n^{2}-81=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 0 for b, and -81 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{0±\sqrt{0^{2}-4\times 9\left(-81\right)}}{2\times 9}$$

Square 0.

$$n=\frac{0±\sqrt{-4\times 9\left(-81\right)}}{2\times 9}$$

Multiply -4 times 9.

$$n=\frac{0±\sqrt{-36\left(-81\right)}}{2\times 9}$$

Multiply -36 times -81.

$$n=\frac{0±\sqrt{2916}}{2\times 9}$$

Take the square root of 2916.

$$n=\frac{0±54}{2\times 9}$$

Multiply 2 times 9.

$$n=\frac{0±54}{18}$$

Now solve the equation n=\frac{0±54}{18} when ± is plus. Divide 54 by 18.

$$n=3$$

Now solve the equation n=\frac{0±54}{18} when ± is minus. Divide -54 by 18.

$$n=-3$$

The equation is now solved.

$$n=3 n=-3$$