Solve 20^2=D^2+14R-2(12)(14)cosx

$$20 ^ { 2 } = D ^ { 2 } + 14 R - 2 ( 12 ) ( 14 ) \cos x$$

Solve for R

$$R=\frac{336\cos(x)-D^{2}+400}{14}$$

Steps for Solving Linear Equation

Calculate 20 to the power of 2 and get 400.

$$400=D^{2}+14R-2\times 12\times 14\cos(x)$$

Multiply 2 and 12 to get 24.

$$400=D^{2}+14R-24\times 14\cos(x)$$

Multiply 24 and 14 to get 336.

$$400=D^{2}+14R-336\cos(x)$$

Swap sides so that all variable terms are on the left hand side.

$$D^{2}+14R-336\cos(x)=400$$

Add 336\cos(x) to both sides.

$$D^{2}+14R=400+336\cos(x)$$

Subtract D^{2} from both sides.

$$14R=400+336\cos(x)-D^{2}$$

The equation is in standard form.

$$14R=336\cos(x)-D^{2}+400$$

Divide both sides by 14.

$$\frac{14R}{14}=\frac{336\cos(x)-D^{2}+400}{14}$$

Dividing by 14 undoes the multiplication by 14.

$$R=\frac{336\cos(x)-D^{2}+400}{14}$$

Solve for D (complex solution)

$$D=-\sqrt{2\left(168\cos(x)-7R+200\right)}$$

$$D=\sqrt{2\left(168\cos(x)-7R+200\right)}$$

Solve for D

$$D=\sqrt{2\left(168\cos(x)-7R+200\right)}$$

$$D=-\sqrt{2\left(168\cos(x)-7R+200\right)}\text{, }R\leq \frac{8\left(21\cos(x)+25\right)}{7}$$