What are the solutions to the equation x^2+(x+2)^2=212 ?

The solutions to the equation x^2+(x+2)^2=212 are x=-1+sqrt(105),x=-1-sqrt(105)

Find the zeros of x^2+(x+2)^2=212

The zeros of x^2+(x+2)^2=212 are x=-1+sqrt(105),x=-1-sqrt(105)

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x^{2} + (x + 2)^{2} = 212

x = - 1 + 105, x = - 1 - 105

Decimal Notation

x = 9.24695 \dots, x = - 11.24695 \dots

Solution steps

x^{2} + (x + 2)^{2} = 212

Expand

x^{2} + (x + 2)^{2} : 2 x^{2} + 4 x + 4

2 x^{2} + 4 x + 4 = 212

Move

212

to the left side

2 x^{2} + 4 x - 208 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 2 ( - 208 )}{2 \cdot 2}

4^{2} - 4 \cdot 2 (- 208) = 4105

x_{1, 2} = \frac{- 4 \pm 4 105}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 4 + 4 105}{2 \cdot 2}, x_{2} = \frac{- 4 - 4 105}{2 \cdot 2}

x = \frac{- 4 + 4 105}{2 \cdot 2} : - 1 + 105

x = \frac{- 4 - 4 105}{2 \cdot 2} : - 1 - 105

The solutions to the quadratic equation are:

x = - 1 + 105, x = - 1 - 105

- 3 x^{2} = 12 x + 12

5 x^{2} + 4 x - 7 = 0

u^{2} + 4 u + 4 = 0

y, x^{2} + (y + \frac{3}{4})^{2} = 1

4 p^{2} + 23 = 7

What are the solutions to the equation x^2+(x+2)^2=212 ?
The solutions to the equation x^2+(x+2)^2=212 are x=-1+sqrt(105),x=-1-sqrt(105)
Find the zeros of x^2+(x+2)^2=212
The zeros of x^2+(x+2)^2=212 are x=-1+sqrt(105),x=-1-sqrt(105)