What are the solutions to the equation 2y^2+10=-7 ?

The solutions to the equation 2y^2+10=-7 are y=sqrt(17/2)i,y=-sqrt(17/2)i

Find the zeros of 2y^2+10=-7

The zeros of 2y^2+10=-7 are y=sqrt(17/2)i,y=-sqrt(17/2)i

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2 y^{2} + 10 = - 7

y = \frac{17}{2} i, y = - \frac{17}{2} i

Solution steps

2 y^{2} + 10 = - 7

Move

10

to the right side

2 y^{2} = - 17

Divide both sides by

2

y^{2} = - \frac{17}{2}

For

x^{2} = f (a)

the solutions are

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

y = - \frac{17}{2}, y = - - \frac{17}{2}

- \frac{17}{2} = \frac{17}{2} i

- - \frac{17}{2} = - \frac{17}{2} i

y = \frac{17}{2} i, y = - \frac{17}{2} i

\frac{1}{2} + \frac{3}{8} - \frac{1}{8}

- 31 - 4 x = - 5 - 5 (1 + 5 x)

x^{2} + 2 x + 9 = 0

(- \frac{27}{8})^{\frac{2}{3}}

\frac{1}{1 + e ^{- 1}}

What are the solutions to the equation 2y^2+10=-7 ?
The solutions to the equation 2y^2+10=-7 are y=sqrt(17/2)i,y=-sqrt(17/2)i
Find the zeros of 2y^2+10=-7
The zeros of 2y^2+10=-7 are y=sqrt(17/2)i,y=-sqrt(17/2)i