(2r+1)(12r^2+5r-3)=0

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Solution for (2r+1)(12r^2+5r-3)=0 equation:


Simplifying
(2r + 1)(12r2 + 5r + -3) = 0

Reorder the terms:
(1 + 2r)(12r2 + 5r + -3) = 0

Reorder the terms:
(1 + 2r)(-3 + 5r + 12r2) = 0

Multiply (1 + 2r) * (-3 + 5r + 12r2)
(1(-3 + 5r + 12r2) + 2r * (-3 + 5r + 12r2)) = 0
((-3 * 1 + 5r * 1 + 12r2 * 1) + 2r * (-3 + 5r + 12r2)) = 0
((-3 + 5r + 12r2) + 2r * (-3 + 5r + 12r2)) = 0
(-3 + 5r + 12r2 + (-3 * 2r + 5r * 2r + 12r2 * 2r)) = 0
(-3 + 5r + 12r2 + (-6r + 10r2 + 24r3)) = 0

Reorder the terms:
(-3 + 5r + -6r + 12r2 + 10r2 + 24r3) = 0

Combine like terms: 5r + -6r = -1r
(-3 + -1r + 12r2 + 10r2 + 24r3) = 0

Combine like terms: 12r2 + 10r2 = 22r2
(-3 + -1r + 22r2 + 24r3) = 0

Solving
-3 + -1r + 22r2 + 24r3 = 0

Solving for variable 'r'.

The solution to this equation could not be determined.

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