help with algebra

Solve for c:
921600 = 518400+c^2-1440 c cos(20°)

921600 = 518400+c^2-1440 c cos(20°) is equivalent to 518400+c^2-1440 c cos(20°) = 921600:
518400+c^2-1440 c cos(20°) = 921600

Subtract 518400 from both sides:
c^2-1440 c cos(20°) = 403200

Add 518400 cos(20°)^2 to both sides:
c^2-1440 c cos(20°)+518400 cos(20°)^2 = 518400 cos(20°)^2+403200

Write the left hand side as a square:
(c-720 cos(20°))^2 = 518400 cos(20°)^2+403200

Take the square root of both sides:
c-720 cos(20°) = sqrt(518400 cos(20°)^2+403200) or c-720 cos(20°) = -sqrt(518400 cos(20°)^2+403200)

Add 720 cos(20°) to both sides:
c = sqrt(518400 cos(20°)^2+403200)+720 cos(20°) or c-720 cos(20°) = -sqrt(518400 cos(20°)^2+403200)

Add 720 cos(20°) to both sides:
Answer: |  c = sqrt(518400 cos(20°)^2+403200)+720 cos(20°)       or      c = 720 cos(20°)-sqrt(518400 cos(20°)^2+403200)

960^2=720^2+c^2−2(720)(c)*cos(20) solve for c with steps

This is a triangle ( cos - rule )

We have $a = 960,\ b = 720,\ A = 20^{\circ}$

1. sin - rule B?

$\begin{array}{rcll} \frac{ \sin{(B)} } {720} &=& \frac{ \sin{ ( 20^{\circ} )} } {960} \\ \sin{(B)} &=& \frac{720}{960} \cdot \sin{(20^{\circ} ) } \\ \sin{(B)} &=& 0.75 \cdot 0.34202014333\\ \sin{(B)} &=& 0.25651510749\\ B &=& \arcsin{ (0.25651510749) } \\ B &=& 14.8633809491^{\circ} \\ \end{array}$

2.  C?

$\begin{array}{rcll} C &=& 180^{\circ} - (A+B) \\ C &=& 180^{\circ} - (20^{\circ}+14.8633809491^{\circ}) \\ C &=& 180^{\circ} - 34.8633809491^{\circ} \\ C &=& 145.136619051^{\circ} \\ \end{array}$

3. sin - rule c?

$\begin{array}{rcll} \frac{c}{ \sin{(C)} } &=& \frac{960}{ \sin{ ( 20^{\circ} )} } \\ c &=& 960 \cdot \frac{\sin{(C)}}{\sin{ ( 20^{\circ} )}} \\ c &=& 960 \cdot \frac{\sin{(145.136619051^{\circ} )}}{\sin{ ( 20^{\circ} )}} \\ c &=& 960 \cdot \frac{0.57162157869}{0.34202014333} \\ c &=& 960 \cdot 1.67130968701 \\ c &=& 1604.45729953 \\ \end{array}$