What are the solutions to the equation 2sqrt(21^2)+4^2=c^2 ?

The solutions to the equation 2sqrt(21^2)+4^2=c^2 are c=sqrt(58),c=-sqrt(58)

Find the zeros of 2sqrt(21^2)+4^2=c^2

The zeros of 2sqrt(21^2)+4^2=c^2 are c=sqrt(58),c=-sqrt(58)

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

c = 58, c = - 58

Decimal Notation

c = 7.61577 \dots, c = - 7.61577 \dots

Solution steps

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

Switch sides

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

Simplify

2 2 1^{2} + 4^{2} : 58

For

x^{2} = f (a)

the solutions are

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

c = 58, c = - 58

x, x^{2} - 2 x + y^{2} - 4 y + 5 = 0

x, (2 x - 1) (3 x + q + 2) = 3 x + q + 2

x (x + 4) = 12126

x^{2} + x - 12 = 0, x^{1} = 3 x^{2} = - 4

x^{2} - 10 x + 4.8 = 0

What are the solutions to the equation 2sqrt(21^2)+4^2=c^2 ?
The solutions to the equation 2sqrt(21^2)+4^2=c^2 are c=sqrt(58),c=-sqrt(58)
Find the zeros of 2sqrt(21^2)+4^2=c^2
The zeros of 2sqrt(21^2)+4^2=c^2 are c=sqrt(58),c=-sqrt(58)