How do you solve $2 c + 14 = 6 - 4 c^{3}$ $2c + 14= 6- 4c^3$ ?

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Answer 1 · 2018-08-02T11:36:31

$4 c^{3} + 2 c + 8$ $4c^3 + 2c + 8$

$c = - 1.12817$ $c = -1.12817$

$2 c + 14 = 6 - 4 c^{3}$ $2c + 14 = 6 - 4c^3$

Collecting like terms;

$2 c + 4 c^{3} + 14 - 6 = 0$ $2c + 4c^3 + 14 - 6 = 0$

Simplifying;

$4 c^{3} + 2 c + 8 = 0 \to Simplified$ $4c^3 + 2c + 8 = 0 -> "Simplified"$

But solving further, we proceed using the method called, W.Lambert Function or Newton Raphson's Method..

$c = -1.12817...$

How do you solve 2c + 14= 6- 4c^32c+14=6−4c32c + 14= 6- 4c^3?