What are the solutions to the equation 7x^2+42=0 ?

The solutions to the equation 7x^2+42=0 are x=sqrt(6)i,x=-sqrt(6)i

Find the zeros of 7x^2+42=0

The zeros of 7x^2+42=0 are x=sqrt(6)i,x=-sqrt(6)i

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7 x^{2} + 42 = 0

x = 6 i, x = - 6 i

Solution steps

7 x^{2} + 42 = 0

Move

42

to the right side

7 x^{2} = - 42

Divide both sides by

7

x^{2} = - 6

For

x^{2} = f (a)

the solutions are

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

x = - 6, x = - - 6

- 6 = 6 i

- - 6 = - 6 i

x = 6 i, x = - 6 i

(6 x - 15 y - 36) \cdot \frac{1}{3}

- x + 2 = 5

2 (x + 1) \leq 0

- 13 < 3 x + 2 \leq 5

7 (p + 3) + 9 = 5 (p - 2) - 3 p

What are the solutions to the equation 7x^2+42=0 ?
The solutions to the equation 7x^2+42=0 are x=sqrt(6)i,x=-sqrt(6)i
Find the zeros of 7x^2+42=0
The zeros of 7x^2+42=0 are x=sqrt(6)i,x=-sqrt(6)i