What are the solutions to the equation x^2-41=-5 ?

The solutions to the equation x^2-41=-5 are x=6,x=-6

Find the zeros of x^2-41=-5

The zeros of x^2-41=-5 are x=6,x=-6

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x^{2} - 41 = - 5

x = 6, x = - 6

Solution steps

x^{2} - 41 = - 5

Move

41

to the right side

x^{2} = 36

For

x^{2} = f (a)

the solutions are

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

x = 36, x = - 36

36 = 6

- 36 = - 6

x = 6, x = - 6

- 45 = 10 t - 4.9 t^{2}

x, y = 4 - (x + 2)^{2}

x^{2} - 9 x - 20 = - 2 x - 2

(x - 36) (x + 36) = 0

x, x^{2} + y^{2} + z^{2} - 2 x + 4 y + 4 = 0

What are the solutions to the equation x^2-41=-5 ?
The solutions to the equation x^2-41=-5 are x=6,x=-6
Find the zeros of x^2-41=-5
The zeros of x^2-41=-5 are x=6,x=-6