What are the solutions to the equation 7^2+4sqrt(2)^2=c^2 ?

The solutions to the equation 7^2+4sqrt(2)^2=c^2 are c=sqrt(57),c=-sqrt(57)

Find the zeros of 7^2+4sqrt(2)^2=c^2

The zeros of 7^2+4sqrt(2)^2=c^2 are c=sqrt(57),c=-sqrt(57)

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7^{2} + 4 2^{2} = c^{2}

c = 57, c = - 57

Decimal Notation

c = 7.54983 \dots, c = - 7.54983 \dots

Solution steps

7^{2} + 4 (2)^{2} = c^{2}

Switch sides

c^{2} = 7^{2} + 4 (2)^{2}

Simplify

7^{2} + 4 (2)^{2} : 57

For

x^{2} = f (a)

the solutions are

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

c = 57, c = - 57

10.4 (t^{2} - 6.2) = 24

4 x^{2} + 20 x + 25 = 0

(2 y - 3)^{2} = 5

3 x^{2} - 14 - 5 = 0

x^{2} - 3 x + 3 = 0

What are the solutions to the equation 7^2+4sqrt(2)^2=c^2 ?
The solutions to the equation 7^2+4sqrt(2)^2=c^2 are c=sqrt(57),c=-sqrt(57)
Find the zeros of 7^2+4sqrt(2)^2=c^2
The zeros of 7^2+4sqrt(2)^2=c^2 are c=sqrt(57),c=-sqrt(57)