What are the solutions to the equation (x+3)^2+(x+2)^2=9 ?

The solutions to the equation (x+3)^2+(x+2)^2=9 are x=(-5+sqrt(17))/2 ,x=-(5+sqrt(17))/2

Find the zeros of (x+3)^2+(x+2)^2=9

The zeros of (x+3)^2+(x+2)^2=9 are x=(-5+sqrt(17))/2 ,x=-(5+sqrt(17))/2

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(x + 3)^{2} + (x + 2)^{2} = 9

x = \frac{- 5 + 17}{2}, x = - \frac{5 + 17}{2}

Solution steps

(x + 3)^{2} + (x + 2)^{2} = 9

Expand

(x + 3)^{2} + (x + 2)^{2} : 2 x^{2} + 10 x + 13

2 x^{2} + 10 x + 13 = 9

Move

9

to the left side

2 x^{2} + 10 x + 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 2 \cdot 4}{2 \cdot 2}

1 0^{2} - 4 \cdot 2 \cdot 4 = 217

x_{1, 2} = \frac{- 10 \pm 2 17}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 10 + 2 17}{2 \cdot 2}, x_{2} = \frac{- 10 - 2 17}{2 \cdot 2}

x = \frac{- 10 + 2 17}{2 \cdot 2} : \frac{- 5 + 17}{2}

x = \frac{- 10 - 2 17}{2 \cdot 2} : - \frac{5 + 17}{2}

The solutions to the quadratic equation are:

x = \frac{- 5 + 17}{2}, x = - \frac{5 + 17}{2}

4 = x^{2} + 10 x

2 x - 5 (x + 3) = x (x - 4)

3 x + 4 = 2 x^{2} - 7

407 x^{2} + 480 x - 69 = 0

(2 x + 5) (6 x - 1) = 120

What are the solutions to the equation (x+3)^2+(x+2)^2=9 ?
The solutions to the equation (x+3)^2+(x+2)^2=9 are x=(-5+sqrt(17))/2 ,x=-(5+sqrt(17))/2
Find the zeros of (x+3)^2+(x+2)^2=9
The zeros of (x+3)^2+(x+2)^2=9 are x=(-5+sqrt(17))/2 ,x=-(5+sqrt(17))/2