What are the solutions to the equation 20^2+b^2=52^2 ?

The solutions to the equation 20^2+b^2=52^2 are b=48,b=-48

Find the zeros of 20^2+b^2=52^2

The zeros of 20^2+b^2=52^2 are b=48,b=-48

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2 0^{2} + b^{2} = 5 2^{2}

b = 48, b = - 48

Solution steps

2 0^{2} + b^{2} = 5 2^{2}

Move

2 0^{2}

to the right side

b^{2} = 2304

For

x^{2} = f (a)

the solutions are

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

b = 2304, b = - 2304

2304 = 48

- 2304 = - 48

b = 48, b = - 48

(x + 1)^{2} = 13

(x + 1)^{2} = 15

2 = 3 x^{2} - x + 2

x^{2} - 2.459 x + 1.5 = 0

2 x^{2} - 6 x + 1 = 0

What are the solutions to the equation 20^2+b^2=52^2 ?
The solutions to the equation 20^2+b^2=52^2 are b=48,b=-48
Find the zeros of 20^2+b^2=52^2
The zeros of 20^2+b^2=52^2 are b=48,b=-48