Solve 4r^2+4gamma+1=0

$$4 r ^ { 2 } + 4 \gamma + 1 = 0$$

Solve for γ

$$\gamma =-r^{2}-\frac{1}{4}$$

Steps for Solving Linear Equation

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

$$4\gamma +1=-4r^{2}$$

Subtract 1 from both sides.

$$4\gamma =-4r^{2}-1$$

Divide both sides by 4.

$$\frac{4\gamma }{4}=\frac{-4r^{2}-1}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$\gamma =\frac{-4r^{2}-1}{4}$$

Divide -4r^{2}-1 by 4.

$$\gamma =-r^{2}-\frac{1}{4}$$

Solve for r (complex solution)

$$r=-\frac{\sqrt{-4\gamma -1}}{2}$$

$$r=\frac{\sqrt{-4\gamma -1}}{2}$$

Solve for r

$$r=\frac{\sqrt{-4\gamma -1}}{2}$$

$$r=-\frac{\sqrt{-4\gamma -1}}{2}\text{, }\gamma \leq -\frac{1}{4}$$