What are the solutions to the equation (3r+1)^2+(r+2)^2=65 ?

The solutions to the equation (3r+1)^2+(r+2)^2=65 are r=2,r=-3

Find the zeros of (3r+1)^2+(r+2)^2=65

The zeros of (3r+1)^2+(r+2)^2=65 are r=2,r=-3

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(3 r + 1)^{2} + (r + 2)^{2} = 65

r = 2, r = - 3

Solution steps

(3 r + 1)^{2} + (r + 2)^{2} = 65

Expand

(3 r + 1)^{2} + (r + 2)^{2} : 10 r^{2} + 10 r + 5

10 r^{2} + 10 r + 5 = 65

Move

65

to the left side

10 r^{2} + 10 r - 60 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 10 ( - 60 )}{2 \cdot 10}

1 0^{2} - 4 \cdot 10 (- 60) = 50

r_{1, 2} = \frac{- 10 \pm 50}{2 \cdot 10}

Separate the solutions

r_{1} = \frac{- 10 + 50}{2 \cdot 10}, r_{2} = \frac{- 10 - 50}{2 \cdot 10}

r = \frac{- 10 + 50}{2 \cdot 10} : 2

r = \frac{- 10 - 50}{2 \cdot 10} : - 3

The solutions to the quadratic equation are:

r = 2, r = - 3

0 = a x^{2} + b x + c

x^{2} + 8 x + 16 = \frac{16}{9}

20 x^{2} - 9 x = 20

5 n^{2} - 20 n + 6 = 0

x, x^{2} - y^{3} = 0

What are the solutions to the equation (3r+1)^2+(r+2)^2=65 ?
The solutions to the equation (3r+1)^2+(r+2)^2=65 are r=2,r=-3
Find the zeros of (3r+1)^2+(r+2)^2=65
The zeros of (3r+1)^2+(r+2)^2=65 are r=2,r=-3