What are the solutions to the equation 3Q^2-192Q+3500=0 ?

The solutions to the equation 3Q^2-192Q+3500=0 are Q=32+i(2sqrt(321))/3 ,Q=32-i(2sqrt(321))/3

Find the zeros of 3Q^2-192Q+3500=0

The zeros of 3Q^2-192Q+3500=0 are Q=32+i(2sqrt(321))/3 ,Q=32-i(2sqrt(321))/3

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3 Q^{2} - 192 Q + 3500 = 0

Q = 32 + i \frac{2 321}{3}, Q = 32 - i \frac{2 321}{3}

Solution steps

3 Q^{2} - 192 Q + 3500 = 0

Solve with the quadratic formula

Q_{1, 2} = \frac{- ( - 192 ) \pm ( - 192 ) ^{2} - 4 \cdot 3 \cdot 3500}{2 \cdot 3}

Simplify

(- 192)^{2} - 4 \cdot 3 \cdot 3500 : 4321 i

Q_{1, 2} = \frac{- ( - 192 ) \pm 4 321 i}{2 \cdot 3}

Separate the solutions

Q_{1} = \frac{- ( - 192 ) + 4 321 i}{2 \cdot 3}, Q_{2} = \frac{- ( - 192 ) - 4 321 i}{2 \cdot 3}

Q = \frac{- ( - 192 ) + 4 321 i}{2 \cdot 3} : 32 + i \frac{2 321}{3}

Q = \frac{- ( - 192 ) - 4 321 i}{2 \cdot 3} : 32 - i \frac{2 321}{3}

The solutions to the quadratic equation are:

Q = 32 + i \frac{2 321}{3}, Q = 32 - i \frac{2 321}{3}

20 x^{2} - 130 x = - 200

3 x^{2} + x + 1

3 x^{2} + x - 2

- 7 x^{2} - 2 x = 1

x^{2} + \frac{x}{20} + \frac{1}{1250} = 0

What are the solutions to the equation 3Q^2-192Q+3500=0 ?
The solutions to the equation 3Q^2-192Q+3500=0 are Q=32+i(2sqrt(321))/3 ,Q=32-i(2sqrt(321))/3
Find the zeros of 3Q^2-192Q+3500=0
The zeros of 3Q^2-192Q+3500=0 are Q=32+i(2sqrt(321))/3 ,Q=32-i(2sqrt(321))/3