What are the solutions to the equation 12-(x-1)^2=x^2+1 ?

The solutions to the equation 12-(x-1)^2=x^2+1 are x=-(-1+sqrt(21))/2 ,x=(1+sqrt(21))/2

Find the zeros of 12-(x-1)^2=x^2+1

The zeros of 12-(x-1)^2=x^2+1 are x=-(-1+sqrt(21))/2 ,x=(1+sqrt(21))/2

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12 - (x - 1)^{2} = x^{2} + 1

x = - \frac{- 1 + 21}{2}, x = \frac{1 + 21}{2}

Decimal Notation

x = - 1.79128 \dots, x = 2.79128 \dots

Solution steps

12 - (x - 1)^{2} = x^{2} + 1

Expand

12 - (x - 1)^{2} : - x^{2} + 2 x + 11

- x^{2} + 2 x + 11 = x^{2} + 1

Move

1

to the left side

- x^{2} + 2 x + 10 = x^{2}

Move

x^{2}

to the left side

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

Solve with the quadratic formula

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

2^{2} - 4 (- 2) \cdot 10 = 221

x_{1, 2} = \frac{- 2 \pm 2 21}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- 2 + 2 21}{2 ( - 2 )}, x_{2} = \frac{- 2 - 2 21}{2 ( - 2 )}

x = \frac{- 2 + 2 21}{2 ( - 2 )} : - \frac{- 1 + 21}{2}

x = \frac{- 2 - 2 21}{2 ( - 2 )} : \frac{1 + 21}{2}

The solutions to the quadratic equation are:

x = - \frac{- 1 + 21}{2}, x = \frac{1 + 21}{2}

10 d^{2} + 15 d - 25 = 6 d^{2}

x, (x - a)^{2} + (y - b)^{2} = r^{2}

x^{2} + x = e

x, 3 x^{2} - 6 x - 3 y^{2} + 3 = 0

x^{2} + 4 x - 65 = 4 x - 1

What are the solutions to the equation 12-(x-1)^2=x^2+1 ?
The solutions to the equation 12-(x-1)^2=x^2+1 are x=-(-1+sqrt(21))/2 ,x=(1+sqrt(21))/2
Find the zeros of 12-(x-1)^2=x^2+1
The zeros of 12-(x-1)^2=x^2+1 are x=-(-1+sqrt(21))/2 ,x=(1+sqrt(21))/2