What are the solutions to the equation c^2=4^2+15^2 ?

The solutions to the equation c^2=4^2+15^2 are c=sqrt(241),c=-sqrt(241)

Find the zeros of c^2=4^2+15^2

The zeros of c^2=4^2+15^2 are c=sqrt(241),c=-sqrt(241)

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c^{2} = 4^{2} + 1 5^{2}

c = 241, c = - 241

Decimal Notation

c = 15.52417 \dots, c = - 15.52417 \dots

Solution steps

c^{2} = 4^{2} + 1 5^{2}

Simplify

4^{2} + 1 5^{2} : 241

For

x^{2} = f (a)

the solutions are

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

c = 241, c = - 241

5 m^{2} - 37 m - 72 = 0

48 + 32 t - 16 t^{2} = 0

0 = \frac{1}{2} x^{2}

k^{2} + 10 k + 10 = 9

c^{2} + 5 c - 24 = 0

What are the solutions to the equation c^2=4^2+15^2 ?
The solutions to the equation c^2=4^2+15^2 are c=sqrt(241),c=-sqrt(241)
Find the zeros of c^2=4^2+15^2
The zeros of c^2=4^2+15^2 are c=sqrt(241),c=-sqrt(241)