Solve 4a_n^2+105n=98

$$4 a _ { n } ^ { 2 } + 105 n = 98$$

Solve for n

$$n=-\frac{4a_{n}^{2}}{105}+\frac{14}{15}$$

Steps for Solving Linear Equation

Subtract 4a_{n}^{2} from both sides.

$$105n=98-4a_{n}^{2}$$

Divide both sides by 105.

$$\frac{105n}{105}=\frac{98-4a_{n}^{2}}{105}$$

Dividing by 105 undoes the multiplication by 105.

$$n=\frac{98-4a_{n}^{2}}{105}$$

Divide 98-4a_{n}^{2} by 105.

$$n=-\frac{4a_{n}^{2}}{105}+\frac{14}{15}$$

Solve for a_n

$$a_{n}=\frac{\sqrt{98-105n}}{2}$$

$$a_{n}=-\frac{\sqrt{98-105n}}{2}\text{, }n\leq \frac{14}{15}$$