What are the solutions to the equation 2x^2+5=-31 ?

The solutions to the equation 2x^2+5=-31 are x=sqrt(2)*3i,x=-sqrt(2)*3i

Find the zeros of 2x^2+5=-31

The zeros of 2x^2+5=-31 are x=sqrt(2)*3i,x=-sqrt(2)*3i

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2 x^{2} + 5 = - 31

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

Solution steps

2 x^{2} + 5 = - 31

Move

5

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

\frac{3}{( 2 ( \frac{1}{2} ) + 3 ) ^{\frac{5}{2}}}

- \frac{3}{4} x^{2} + 2 = 0

- 10 y^{4} (4 x y^{2})

3 + 7 (- 3 - 2 r) = - 102

4 x^{2} - 4 x - 3

What are the solutions to the equation 2x^2+5=-31 ?
The solutions to the equation 2x^2+5=-31 are x=sqrt(2)*3i,x=-sqrt(2)*3i
Find the zeros of 2x^2+5=-31
The zeros of 2x^2+5=-31 are x=sqrt(2)*3i,x=-sqrt(2)*3i