What are the solutions to the equation 14^2+20^2=c^2 ?

The solutions to the equation 14^2+20^2=c^2 are c=2sqrt(149),c=-2sqrt(149)

Find the zeros of 14^2+20^2=c^2

The zeros of 14^2+20^2=c^2 are c=2sqrt(149),c=-2sqrt(149)

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1 4^{2} + 2 0^{2} = c^{2}

c = 2149, c = - 2149

Decimal Notation

c = 24.41311 \dots, c = - 24.41311 \dots

Solution steps

1 4^{2} + 2 0^{2} = c^{2}

Switch sides

c^{2} = 1 4^{2} + 2 0^{2}

Simplify

1 4^{2} + 2 0^{2} : 596

For

x^{2} = f (a)

the solutions are

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

c = 596, c = - 596

596 = 2149

- 596 = - 2149

c = 2149, c = - 2149

x^{2} + 12 x + 400 = 0

x^{2} + 5 x + 1 = 8

6 p^{2} - 5 = 2 + 5 p^{2} - 6 p

0 = 3 t^{2} - 12 t - 15

4 x^{2} + 180 = 0

What are the solutions to the equation 14^2+20^2=c^2 ?
The solutions to the equation 14^2+20^2=c^2 are c=2sqrt(149),c=-2sqrt(149)
Find the zeros of 14^2+20^2=c^2
The zeros of 14^2+20^2=c^2 are c=2sqrt(149),c=-2sqrt(149)