(3r+2)(2r^2+4r-3)=0

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Solution for (3r+2)(2r^2+4r-3)=0 equation:


Simplifying
(3r + 2)(2r2 + 4r + -3) = 0

Reorder the terms:
(2 + 3r)(2r2 + 4r + -3) = 0

Reorder the terms:
(2 + 3r)(-3 + 4r + 2r2) = 0

Multiply (2 + 3r) * (-3 + 4r + 2r2)
(2(-3 + 4r + 2r2) + 3r * (-3 + 4r + 2r2)) = 0
((-3 * 2 + 4r * 2 + 2r2 * 2) + 3r * (-3 + 4r + 2r2)) = 0
((-6 + 8r + 4r2) + 3r * (-3 + 4r + 2r2)) = 0
(-6 + 8r + 4r2 + (-3 * 3r + 4r * 3r + 2r2 * 3r)) = 0
(-6 + 8r + 4r2 + (-9r + 12r2 + 6r3)) = 0

Reorder the terms:
(-6 + 8r + -9r + 4r2 + 12r2 + 6r3) = 0

Combine like terms: 8r + -9r = -1r
(-6 + -1r + 4r2 + 12r2 + 6r3) = 0

Combine like terms: 4r2 + 12r2 = 16r2
(-6 + -1r + 16r2 + 6r3) = 0

Solving
-6 + -1r + 16r2 + 6r3 = 0

Solving for variable 'r'.

The solution to this equation could not be determined.

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