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

The solutions to the equation 2x^2+6x=1 are x=(-3+sqrt(11))/2 ,x=-(3+sqrt(11))/2

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

The zeros of 2x^2+6x=1 are x=(-3+sqrt(11))/2 ,x=-(3+sqrt(11))/2

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2 x^{2} + 6 x = 1

x = \frac{- 3 + 11}{2}, x = - \frac{3 + 11}{2}

Decimal Notation

x = 0.15831 \dots, x = - 3.15831 \dots

Solution steps

2 x^{2} + 6 x = 1

Move

1

to the left side

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

Solve with the quadratic formula

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

6^{2} - 4 \cdot 2 (- 1) = 211

x_{1, 2} = \frac{- 6 \pm 2 11}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 6 + 2 11}{2 \cdot 2}, x_{2} = \frac{- 6 - 2 11}{2 \cdot 2}

x = \frac{- 6 + 2 11}{2 \cdot 2} : \frac{- 3 + 11}{2}

x = \frac{- 6 - 2 11}{2 \cdot 2} : - \frac{3 + 11}{2}

The solutions to the quadratic equation are:

x = \frac{- 3 + 11}{2}, x = - \frac{3 + 11}{2}

2 x^{2} + x - 6 \geq 0

(\frac{4}{9})^{2}

\frac{x ^{2}}{18} - \frac{x}{9} = 1

3 x + 12 = \frac{x + 7}{3}

i^{89}

What are the solutions to the equation 2x^2+6x=1 ?
The solutions to the equation 2x^2+6x=1 are x=(-3+sqrt(11))/2 ,x=-(3+sqrt(11))/2
Find the zeros of 2x^2+6x=1
The zeros of 2x^2+6x=1 are x=(-3+sqrt(11))/2 ,x=-(3+sqrt(11))/2