Solve 5R_2^2+6R_{1}-19R_{2}=0

$$5 R _ { 2 } ^ { 2 } + 6 R _ { 1 } - 19 R _ { 2 } = 0$$

Solve for R_1

$$R_{1}=-\frac{R_{2}\left(5R_{2}-19\right)}{6}$$

Steps for Solving Linear Equation

Subtract 5R_{2}^{2} from both sides. Anything subtracted from zero gives its negation.

$$6R_{1}-19R_{2}=-5R_{2}^{2}$$

Add 19R_{2} to both sides.

$$6R_{1}=-5R_{2}^{2}+19R_{2}$$

The equation is in standard form.

$$6R_{1}=19R_{2}-5R_{2}^{2}$$

Divide both sides by 6.

$$\frac{6R_{1}}{6}=\frac{R_{2}\left(19-5R_{2}\right)}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$R_{1}=\frac{R_{2}\left(19-5R_{2}\right)}{6}$$

Solve for R_2

$$R_{2}=\frac{\sqrt{361-120R_{1}}+19}{10}$$

$$R_{2}=\frac{-\sqrt{361-120R_{1}}+19}{10}\text{, }R_{1}\leq \frac{361}{120}$$