What are the solutions to the equation x^2+6x+3=1 ?

The solutions to the equation x^2+6x+3=1 are x=-3+sqrt(7),x=-3-sqrt(7)

Find the zeros of x^2+6x+3=1

The zeros of x^2+6x+3=1 are x=-3+sqrt(7),x=-3-sqrt(7)

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x^{2} + 6 x + 3 = 1

x = - 3 + 7, x = - 3 - 7

Decimal Notation

x = - 0.35424 \dots, x = - 5.64575 \dots

Solution steps

x^{2} + 6 x + 3 = 1

Move

1

to the left side

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

Solve with the quadratic formula

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

6^{2} - 4 \cdot 1 \cdot 2 = 27

x_{1, 2} = \frac{- 6 \pm 2 7}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 6 + 2 7}{2 \cdot 1}, x_{2} = \frac{- 6 - 2 7}{2 \cdot 1}

x = \frac{- 6 + 2 7}{2 \cdot 1} : - 3 + 7

x = \frac{- 6 - 2 7}{2 \cdot 1} : - 3 - 7

The solutions to the quadratic equation are:

x = - 3 + 7, x = - 3 - 7

x^{2} = 40 - 18 x

3 x - 2 = - 9 x^{2}

3 b^{2} + 3 b = 6

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

4 x^{2} - 17 x = 15

What are the solutions to the equation x^2+6x+3=1 ?
The solutions to the equation x^2+6x+3=1 are x=-3+sqrt(7),x=-3-sqrt(7)
Find the zeros of x^2+6x+3=1
The zeros of x^2+6x+3=1 are x=-3+sqrt(7),x=-3-sqrt(7)