What are the solutions to the equation x^2+(x+3)^2=29 ?

The solutions to the equation x^2+(x+3)^2=29 are x=2,x=-5

Find the zeros of x^2+(x+3)^2=29

The zeros of x^2+(x+3)^2=29 are x=2,x=-5

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x^{2} + (x + 3)^{2} = 29

x = 2, x = - 5

Solution steps

x^{2} + (x + 3)^{2} = 29

Expand

x^{2} + (x + 3)^{2} : 2 x^{2} + 6 x + 9

2 x^{2} + 6 x + 9 = 29

Move

29

to the left side

2 x^{2} + 6 x - 20 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 2 ( - 20 )}{2 \cdot 2}

6^{2} - 4 \cdot 2 (- 20) = 14

x_{1, 2} = \frac{- 6 \pm 14}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 6 + 14}{2 \cdot 2}, x_{2} = \frac{- 6 - 14}{2 \cdot 2}

x = \frac{- 6 + 14}{2 \cdot 2} : 2

x = \frac{- 6 - 14}{2 \cdot 2} : - 5

The solutions to the quadratic equation are:

x = 2, x = - 5

0.548983 = [1 - (\frac{1 - c}{0.2})^{2}]

x^{2} - 9 x - 1 = 9

2 x^{2} + 3 x + 11 = 0

x^{2} + 10 x + 12

(x - 12)^{2} = 64

What are the solutions to the equation x^2+(x+3)^2=29 ?
The solutions to the equation x^2+(x+3)^2=29 are x=2,x=-5
Find the zeros of x^2+(x+3)^2=29
The zeros of x^2+(x+3)^2=29 are x=2,x=-5