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

The solutions to the equation x^2+6x+1=-2x+1-2x+1 are x=-5+sqrt(26),x=-5-sqrt(26)

Find the zeros of x^2+6x+1=-2x+1-2x+1

The zeros of x^2+6x+1=-2x+1-2x+1 are x=-5+sqrt(26),x=-5-sqrt(26)

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

x = - 5 + 26, x = - 5 - 26

Decimal Notation

x = 0.09901 \dots, x = - 10.09901 \dots

Solution steps

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

Refine

x^{2} + 6 x + 1 = - 4 x + 2

Move

2

to the left side

x^{2} + 6 x - 1 = - 4 x

Move

4 x

to the left side

x^{2} + 10 x - 1 = 0

Solve with the quadratic formula

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

1 0^{2} - 4 \cdot 1 \cdot (- 1) = 226

x_{1, 2} = \frac{- 10 \pm 2 26}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 10 + 2 26}{2 \cdot 1}, x_{2} = \frac{- 10 - 2 26}{2 \cdot 1}

x = \frac{- 10 + 2 26}{2 \cdot 1} : - 5 + 26

x = \frac{- 10 - 2 26}{2 \cdot 1} : - 5 - 26

The solutions to the quadratic equation are:

x = - 5 + 26, x = - 5 - 26

x^{2} - 81 = (x - 9) 2

x, x^{2} - 4 y^{2} - 8 y = 4

x, 2 x^{2} + 5 y^{2} + z^{2} = 4

u^{2} - 8 u = 0

5 X^{2} - 2 X - 1 = 0

What are the solutions to the equation x^2+6x+1=-2x+1-2x+1 ?
The solutions to the equation x^2+6x+1=-2x+1-2x+1 are x=-5+sqrt(26),x=-5-sqrt(26)
Find the zeros of x^2+6x+1=-2x+1-2x+1
The zeros of x^2+6x+1=-2x+1-2x+1 are x=-5+sqrt(26),x=-5-sqrt(26)