What are the solutions to the equation (r+5)^2+(r-2)^2=37 ?

The solutions to the equation (r+5)^2+(r-2)^2=37 are r=1,r=-4

Find the zeros of (r+5)^2+(r-2)^2=37

The zeros of (r+5)^2+(r-2)^2=37 are r=1,r=-4

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(r + 5)^{2} + (r - 2)^{2} = 37

r = 1, r = - 4

Solution steps

(r + 5)^{2} + (r - 2)^{2} = 37

Expand

(r + 5)^{2} + (r - 2)^{2} : 2 r^{2} + 6 r + 29

2 r^{2} + 6 r + 29 = 37

Move

37

to the left side

2 r^{2} + 6 r - 8 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 2 ( - 8 )}{2 \cdot 2}

6^{2} - 4 \cdot 2 (- 8) = 10

r_{1, 2} = \frac{- 6 \pm 10}{2 \cdot 2}

Separate the solutions

r_{1} = \frac{- 6 + 10}{2 \cdot 2}, r_{2} = \frac{- 6 - 10}{2 \cdot 2}

r = \frac{- 6 + 10}{2 \cdot 2} : 1

r = \frac{- 6 - 10}{2 \cdot 2} : - 4

The solutions to the quadratic equation are:

r = 1, r = - 4

24 x^{2} + 114 x + 132 = 0

x^{2} + 4^{2} = 8^{2}

(x + 4)^{2} = 18

- x^{2} - 4 x + 21 = 0

10 x^{2} + 5 x + 1 = 2

What are the solutions to the equation (r+5)^2+(r-2)^2=37 ?
The solutions to the equation (r+5)^2+(r-2)^2=37 are r=1,r=-4
Find the zeros of (r+5)^2+(r-2)^2=37
The zeros of (r+5)^2+(r-2)^2=37 are r=1,r=-4