Solve 7(2n-3) = 21n^21

Question

Solve the equation(The real numbers system)

n \in / R

Alternative Form

No real solution

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Solve using the quadratic formula in the complex numbers system

Solve by completing the square in the complex numbers system

Solve using the PQ formula in the complex numbers system

n_{1} = \frac{1}{3} - \frac{2 2}{3} i, n_{2} = \frac{1}{3} + \frac{2 2}{3} i

Alternative Form

n_{1} \approx 0. \dot{3} - 0.942809 i, n_{2} \approx 0. \dot{3} + 0.942809 i

Evaluate

7 (2 n - 3) = 21 n^{2} \times 1

Multiply the terms

7 (2 n - 3) = 21 n^{2}

Swap the sides

21 n^{2} = 7 (2 n - 3)

Expand the expression

More Steps

21 n^{2} = 14 n - 21

Move the expression to the left side

21 n^{2} - 14 n + 21 = 0

Substitute a= 21,b= - 14 and c= 21 into the quadratic formula n = \frac{- b \pm b ^{2} - 4 a c}{2 a}

n = \frac{14 \pm ( - 14 ) ^{2} - 4 \times 21 \times 21}{2 \times 21}

Simplify the expression

n = \frac{14 \pm ( - 14 ) ^{2} - 4 \times 21 \times 21}{42}

Simplify the expression

More Steps

n = \frac{14 \pm - 1568}{42}

Simplify the radical expression

More Steps

n = \frac{14 \pm 28 2 \times i}{42}

Separate the equation into 2 possible cases

n = \frac{14 + 28 2 \times i}{42} n = \frac{14 - 28 2 \times i}{42}

Simplify the expression

More Steps

n = \frac{1}{3} + \frac{2 2}{3} i n = \frac{14 - 28 2 \times i}{42}

Simplify the expression

More Steps

n = \frac{1}{3} + \frac{2 2}{3} i n = \frac{1}{3} - \frac{2 2}{3} i

Solution

n_{1} = \frac{1}{3} - \frac{2 2}{3} i, n_{2} = \frac{1}{3} + \frac{2 2}{3} i

Alternative Form

n_{1} \approx 0. \dot{3} - 0.942809 i, n_{2} \approx 0. \dot{3} + 0.942809 i

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