Solve 16k^2-4R+13=0

$$16 { k }^{ 2 } -4R+13 = 0$$

Solve for R

$$R=4k^{2}+\frac{13}{4}$$

Steps for Solving Linear Equation

Subtract 16k^{2} from both sides. Anything subtracted from zero gives its negation.

$$-4R+13=-16k^{2}$$

Subtract 13 from both sides.

$$-4R=-16k^{2}-13$$

Divide both sides by -4.

$$\frac{-4R}{-4}=\frac{-16k^{2}-13}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$R=\frac{-16k^{2}-13}{-4}$$

Divide -16k^{2}-13 by -4.

$$R=4k^{2}+\frac{13}{4}$$

Solve for k (complex solution)

$$k=-\frac{\sqrt{4R-13}}{4}$$

$$k=\frac{\sqrt{4R-13}}{4}$$

Solve for k

$$k=\frac{\sqrt{4R-13}}{4}$$

$$k=-\frac{\sqrt{4R-13}}{4}\text{, }R\geq \frac{13}{4}$$