What are the solutions to the equation 21^2+28^2=c^2 ?

The solutions to the equation 21^2+28^2=c^2 are c=35,c=-35

Find the zeros of 21^2+28^2=c^2

The zeros of 21^2+28^2=c^2 are c=35,c=-35

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2 1^{2} + 2 8^{2} = c^{2}

c = 35, c = - 35

Solution steps

2 1^{2} + 2 8^{2} = c^{2}

Switch sides

c^{2} = 2 1^{2} + 2 8^{2}

Simplify

2 1^{2} + 2 8^{2} : 1225

For

x^{2} = f (a)

the solutions are

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

c = 1225, c = - 1225

1225 = 35

- 1225 = - 35

c = 35, c = - 35

0 = 48 x + 8 x^{2}

x^{2} + 2 (1)^{2} - 2 (1) - 5 = 16

x^{2} - 6 x - 4 = 0

a^{2} + 1 = 3 a - 1

3 + 9 r^{2} = 12

What are the solutions to the equation 21^2+28^2=c^2 ?
The solutions to the equation 21^2+28^2=c^2 are c=35,c=-35
Find the zeros of 21^2+28^2=c^2
The zeros of 21^2+28^2=c^2 are c=35,c=-35