What are the solutions to the equation 2x^2+16=-20 ?

The solutions to the equation 2x^2+16=-20 are x=sqrt(2)*3i,x=-sqrt(2)*3i

Find the zeros of 2x^2+16=-20

The zeros of 2x^2+16=-20 are x=sqrt(2)*3i,x=-sqrt(2)*3i

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2 x^{2} + 16 = - 20

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

Solution steps

2 x^{2} + 16 = - 20

Move

16

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

x^{2} - 8 x = 242

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

12 a^{2} b x - 15 a^{2} b y

x - 4 > 1

\frac{( x + 4 ) ( x - 5 )}{( x + 6 ) ( 3 x - 2 )} \geq 0

What are the solutions to the equation 2x^2+16=-20 ?
The solutions to the equation 2x^2+16=-20 are x=sqrt(2)*3i,x=-sqrt(2)*3i
Find the zeros of 2x^2+16=-20
The zeros of 2x^2+16=-20 are x=sqrt(2)*3i,x=-sqrt(2)*3i