What are the solutions to the equation (x+1)^2+(x+2)^2=x^2+6 ?

The solutions to the equation (x+1)^2+(x+2)^2=x^2+6 are x=-3+sqrt(10),x=-3-sqrt(10)

Find the zeros of (x+1)^2+(x+2)^2=x^2+6

The zeros of (x+1)^2+(x+2)^2=x^2+6 are x=-3+sqrt(10),x=-3-sqrt(10)

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(x + 1)^{2} + (x + 2)^{2} = x^{2} + 6

x = - 3 + 10, x = - 3 - 10

Decimal Notation

x = 0.16227 \dots, x = - 6.16227 \dots

Solution steps

(x + 1)^{2} + (x + 2)^{2} = x^{2} + 6

Expand

(x + 1)^{2} + (x + 2)^{2} : 2 x^{2} + 6 x + 5

2 x^{2} + 6 x + 5 = x^{2} + 6

Move

6

to the left side

2 x^{2} + 6 x - 1 = x^{2}

Move

x^{2}

to the left side

x^{2} + 6 x - 1 = 0

Solve with the quadratic formula

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

6^{2} - 4 \cdot 1 \cdot (- 1) = 210

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

Separate the solutions

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

x = \frac{- 6 + 2 10}{2 \cdot 1} : - 3 + 10

x = \frac{- 6 - 2 10}{2 \cdot 1} : - 3 - 10

The solutions to the quadratic equation are:

x = - 3 + 10, x = - 3 - 10

28 r^{2} - 49 r = 21 r

(m - 2) x^{2} + (2 m + 1) x = m^{2} + 5 m + 1

x^{2} + 9 x - 136 = 0

4 d^{2} + 9 d - 2 = 4 d

9 x^{2} - 169 = 0

What are the solutions to the equation (x+1)^2+(x+2)^2=x^2+6 ?
The solutions to the equation (x+1)^2+(x+2)^2=x^2+6 are x=-3+sqrt(10),x=-3-sqrt(10)
Find the zeros of (x+1)^2+(x+2)^2=x^2+6
The zeros of (x+1)^2+(x+2)^2=x^2+6 are x=-3+sqrt(10),x=-3-sqrt(10)