What are the solutions to the equation y^2+22=38 ?

The solutions to the equation y^2+22=38 are y=4,y=-4

Find the zeros of y^2+22=38

The zeros of y^2+22=38 are y=4,y=-4

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y^{2} + 22 = 38

y = 4, y = - 4

Solution steps

y^{2} + 22 = 38

Move

22

to the right side

y^{2} = 16

For

x^{2} = f (a)

the solutions are

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

y = 16, y = - 16

16 = 4

- 16 = - 4

y = 4, y = - 4

4 x^{2} = 2 x + 11

5^{\frac{2}{3}} \cdot 5^{\frac{4}{3}}

3 = \frac{2 x + 19}{5}

m^{2} + 16 m + 40 = 0

y < 15 - 13

What are the solutions to the equation y^2+22=38 ?
The solutions to the equation y^2+22=38 are y=4,y=-4
Find the zeros of y^2+22=38
The zeros of y^2+22=38 are y=4,y=-4