What are the solutions to the equation 2z^2-22=0 ?

The solutions to the equation 2z^2-22=0 are z=sqrt(11),z=-sqrt(11)

Find the zeros of 2z^2-22=0

The zeros of 2z^2-22=0 are z=sqrt(11),z=-sqrt(11)

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2 z^{2} - 22 = 0

z = 11, z = - 11

Solution steps

2 z^{2} - 22 = 0

Move

22

to the right side

2 z^{2} = 22

Divide both sides by

2

z^{2} = 11

For

x^{2} = f (a)

the solutions are

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

z = 11, z = - 11

6 (x - 5) (x - 4) < 0

6 x + 5 < - 25

\frac{( x + 5 ) ( 2 x - 7 )}{( x + 1 ) ^{2}} \leq 0

u^{2} - 10 u + 25

5 t + 30

What are the solutions to the equation 2z^2-22=0 ?
The solutions to the equation 2z^2-22=0 are z=sqrt(11),z=-sqrt(11)
Find the zeros of 2z^2-22=0
The zeros of 2z^2-22=0 are z=sqrt(11),z=-sqrt(11)