What are the solutions to the equation 2x^2+50=4 ?

The solutions to the equation 2x^2+50=4 are x=sqrt(23)i,x=-sqrt(23)i

Find the zeros of 2x^2+50=4

The zeros of 2x^2+50=4 are x=sqrt(23)i,x=-sqrt(23)i

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2 x^{2} + 50 = 4

x = 23 i, x = - 23 i

Solution steps

2 x^{2} + 50 = 4

Move

50

to the right side

2 x^{2} = - 46

Divide both sides by

2

x^{2} = - 23

For

x^{2} = f (a)

the solutions are

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

x = - 23, x = - - 23

- 23 = 23 i

- - 23 = - 23 i

x = 23 i, x = - 23 i

x^{2} = - 9 x

3 6^{\frac{1}{4}} \cdot 3 6^{\frac{1}{4}}

∣ 3 - 2 x ∣ = ∣ x + 5 ∣

2^{3} - [(4) (\frac{6}{3} - 1)]

3 k - 5 = 7 k + 7

What are the solutions to the equation 2x^2+50=4 ?
The solutions to the equation 2x^2+50=4 are x=sqrt(23)i,x=-sqrt(23)i
Find the zeros of 2x^2+50=4
The zeros of 2x^2+50=4 are x=sqrt(23)i,x=-sqrt(23)i