What are the solutions to the equation 12=x^2-4 ?

The solutions to the equation 12=x^2-4 are x=4,x=-4

Find the zeros of 12=x^2-4

The zeros of 12=x^2-4 are x=4,x=-4

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12 = x^{2} - 4

x = 4, x = - 4

Solution steps

12 = x^{2} - 4

Switch sides

x^{2} - 4 = 12

Move

4

to the right side

x^{2} = 16

For

x^{2} = f (a)

the solutions are

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

x = 16, x = - 16

16 = 4

- 16 = - 4

x = 4, x = - 4

z^{2} + (3 + j) z + 1 = 0

2 x^{2} + 6 x = 36

12 x^{2} + 15 x = 0

(a + 1) x^{2} + (2 a + 1) x - a = 30

x, x^{2} + x - 1 = 0

What are the solutions to the equation 12=x^2-4 ?
The solutions to the equation 12=x^2-4 are x=4,x=-4
Find the zeros of 12=x^2-4
The zeros of 12=x^2-4 are x=4,x=-4