What are the solutions to the equation 2^2+(-3^2)=c^2 ?

The solutions to the equation 2^2+(-3^2)=c^2 are c=sqrt(5)i,c=-sqrt(5)i

Find the zeros of 2^2+(-3^2)=c^2

The zeros of 2^2+(-3^2)=c^2 are c=sqrt(5)i,c=-sqrt(5)i

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2^{2} + (- 3^{2}) = c^{2}

c = 5 i, c = - 5 i

Solution steps

2^{2} + (- 3^{2}) = c^{2}

Switch sides

c^{2} = 2^{2} + (- 3^{2})

Simplify

2^{2} + (- 3^{2}) : - 5

For

x^{2} = f (a)

the solutions are

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

c = - 5, c = - - 5

- 5 = 5 i

- - 5 = - 5 i

c = 5 i, c = - 5 i

1 0^{2} + y^{2} = 9^{2}

12 - 8 x + x^{2} = 0

4 x^{2} - 10 x = 2 x^{2} + 5 x

0 = x^{2} - x - 20

(x + 2)^{2} - 3 x = 24

What are the solutions to the equation 2^2+(-3^2)=c^2 ?
The solutions to the equation 2^2+(-3^2)=c^2 are c=sqrt(5)i,c=-sqrt(5)i
Find the zeros of 2^2+(-3^2)=c^2
The zeros of 2^2+(-3^2)=c^2 are c=sqrt(5)i,c=-sqrt(5)i