Solve 4r^2-20gamma+25=0

$$4 r ^ { 2 } - 20 \gamma + 25 = 0$$

Solve for γ

$$\gamma =\frac{r^{2}}{5}+\frac{5}{4}$$

Steps for Solving Linear Equation

Subtract 4r^{2} from both sides. Anything subtracted from zero gives its negation.

$$-20\gamma +25=-4r^{2}$$

Subtract 25 from both sides.

$$-20\gamma =-4r^{2}-25$$

Divide both sides by -20.

$$\frac{-20\gamma }{-20}=\frac{-4r^{2}-25}{-20}$$

Dividing by -20 undoes the multiplication by -20.

$$\gamma =\frac{-4r^{2}-25}{-20}$$

Divide -4r^{2}-25 by -20.

$$\gamma =\frac{r^{2}}{5}+\frac{5}{4}$$

Solve for r

$$r=\frac{\sqrt{20\gamma -25}}{2}$$

$$r=-\frac{\sqrt{20\gamma -25}}{2}\text{, }\gamma \geq \frac{5}{4}$$