What are the solutions to the equation 40(-3x^2-4)=0 ?

The solutions to the equation 40(-3x^2-4)=0 are x= 2/(sqrt(3))i,x=-2/(sqrt(3))i

Find the zeros of 40(-3x^2-4)=0

The zeros of 40(-3x^2-4)=0 are x= 2/(sqrt(3))i,x=-2/(sqrt(3))i

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40 (- 3 x^{2} - 4) = 0

x = \frac{2}{3} i, x = - \frac{2}{3} i

Solution steps

40 (- 3 x^{2} - 4) = 0

Divide both sides by

40

- 3 x^{2} - 4 = 0

Move

4

to the right side

- 3 x^{2} = 4

Divide both sides by

- 3

x^{2} = - \frac{4}{3}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

x = - \frac{4}{3}, x = - - \frac{4}{3}

- \frac{4}{3} = \frac{2}{3} i

- - \frac{4}{3} = - \frac{2}{3} i

x = \frac{2}{3} i, x = - \frac{2}{3} i

r^{2} - 7 r + 10 = 0

- 4 x^{2} + 2 x + 4 = 0

240 x - 3 x^{2} = 0

4 (5 m + 3)^{2} - 28 = 0

C = 400 - 5 x + 0.125 x^{2}

What are the solutions to the equation 40(-3x^2-4)=0 ?
The solutions to the equation 40(-3x^2-4)=0 are x= 2/(sqrt(3))i,x=-2/(sqrt(3))i
Find the zeros of 40(-3x^2-4)=0
The zeros of 40(-3x^2-4)=0 are x= 2/(sqrt(3))i,x=-2/(sqrt(3))i