What are the solutions to the equation 29=x^2+4 ?

The solutions to the equation 29=x^2+4 are x=5,x=-5

Find the zeros of 29=x^2+4

The zeros of 29=x^2+4 are x=5,x=-5

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29 = x^{2} + 4

x = 5, x = - 5

Solution steps

29 = x^{2} + 4

Switch sides

x^{2} + 4 = 29

Move

4

to the right side

x^{2} = 25

For

x^{2} = f (a)

the solutions are

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

x = 25, x = - 25

25 = 5

- 25 = - 5

x = 5, x = - 5

x, f = u_{1} (x^{2})

s^{2} + 4 s - 21 = 0

0 = 75 + 28925 \cdot t - 5 \cdot t^{2}

4 n^{2} - 7 n - 85 = 7

2 x^{2} + 2 x - 6 = x^{2}

What are the solutions to the equation 29=x^2+4 ?
The solutions to the equation 29=x^2+4 are x=5,x=-5
Find the zeros of 29=x^2+4
The zeros of 29=x^2+4 are x=5,x=-5