What are the solutions to the equation 20^2+48^2=c^2 ?

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

Find the zeros of 20^2+48^2=c^2

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

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2 0^{2} + 4 8^{2} = c^{2}

c = 52, c = - 52

Solution steps

2 0^{2} + 4 8^{2} = c^{2}

Switch sides

c^{2} = 2 0^{2} + 4 8^{2}

Simplify

2 0^{2} + 4 8^{2} : 2704

For

x^{2} = f (a)

the solutions are

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

c = 2704, c = - 2704

2704 = 52

- 2704 = - 52

c = 52, c = - 52

x^{2} - (1 + a) x + a = 0

19 x^{2} = 19 x

x^{2} - \frac{x}{2} = \frac{3}{10}

4 x^{2} + 8 x - 2 = 0

(2 s - 1)^{2} - 225 = 0

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