What are the solutions to the equation 3(2x^2+7)=-87 ?

The solutions to the equation 3(2x^2+7)=-87 are x=sqrt(2)*3i,x=-sqrt(2)*3i

Find the zeros of 3(2x^2+7)=-87

The zeros of 3(2x^2+7)=-87 are x=sqrt(2)*3i,x=-sqrt(2)*3i

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3 (2 x^{2} + 7) = - 87

x = 2 \cdot 3 i, x = - 2 \cdot 3 i

Solution steps

3 (2 x^{2} + 7) = - 87

Divide both sides by

3

2 x^{2} + 7 = - 29

Move

7

to the right side

2 x^{2} = - 36

Divide both sides by

2

x^{2} = - 18

For

x^{2} = f (a)

the solutions are

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

x = - 18, x = - - 18

- 18 = 2 \cdot 3 i

- - 18 = - 2 \cdot 3 i

x = 2 \cdot 3 i, x = - 2 \cdot 3 i

y^{2} - 7 y = 0

d^{2} - 5 d - 14 = 0

x^{2} - 6 x + 16

x^{2} + 6 x - 3

w^{2} - 14 w + 42 = 0

What are the solutions to the equation 3(2x^2+7)=-87 ?
The solutions to the equation 3(2x^2+7)=-87 are x=sqrt(2)*3i,x=-sqrt(2)*3i
Find the zeros of 3(2x^2+7)=-87
The zeros of 3(2x^2+7)=-87 are x=sqrt(2)*3i,x=-sqrt(2)*3i