What are the solutions to the equation (2-(4))^2+(1-(-3))^2=r^2 ?

The solutions to the equation (2-(4))^2+(1-(-3))^2=r^2 are r=2sqrt(5),r=-2sqrt(5)

Find the zeros of (2-(4))^2+(1-(-3))^2=r^2

The zeros of (2-(4))^2+(1-(-3))^2=r^2 are r=2sqrt(5),r=-2sqrt(5)

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(2 - (4))^{2} + (1 - (- 3))^{2} = r^{2}

r = 25, r = - 25

Decimal Notation

r = 4.47213 \dots, r = - 4.47213 \dots

Solution steps

(2 - (4))^{2} + (1 - (- 3))^{2} = r^{2}

Switch sides

r^{2} = (2 - (4))^{2} + (1 - (- 3))^{2}

Simplify

(2 - (4))^{2} + (1 - (- 3))^{2} : 20

For

x^{2} = f (a)

the solutions are

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

r = 20, r = - 20

20 = 25

- 20 = - 25

r = 25, r = - 25

2 x^{2} + 17 x + 24 = 0

x^{2} - \frac{1}{4} x + 2

4 (6 - 4 x)^{2} = 49

y^{2} + 3 y - 28 = 0

x^{2} = \frac{1}{16}

What are the solutions to the equation (2-(4))^2+(1-(-3))^2=r^2 ?
The solutions to the equation (2-(4))^2+(1-(-3))^2=r^2 are r=2sqrt(5),r=-2sqrt(5)
Find the zeros of (2-(4))^2+(1-(-3))^2=r^2
The zeros of (2-(4))^2+(1-(-3))^2=r^2 are r=2sqrt(5),r=-2sqrt(5)