Mathematics
Mathematics, 12.03.2020 18:31, filzaf9632

From the previous step we have erx(6r2 + 3r βˆ’ 3) = 0. Since erx is never 0, then we know 6r2 + 3r βˆ’ 3 = 0. Solving the above equation gives the solutions to the quadratic equation.

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From the previous step we have erx(6r2 + 3r βˆ’ 3) = 0. Since erx is never 0, then we know 6r2 + 3r βˆ’...

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