What are the solutions to the equation 3x^2+7x=1 ?

The solutions to the equation 3x^2+7x=1 are x=(-7+sqrt(61))/6 ,x=(-7-sqrt(61))/6

Find the zeros of 3x^2+7x=1

The zeros of 3x^2+7x=1 are x=(-7+sqrt(61))/6 ,x=(-7-sqrt(61))/6

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3 x^{2} + 7 x = 1

x = \frac{- 7 + 61}{6}, x = \frac{- 7 - 61}{6}

Decimal Notation

x = 0.13504 \dots, x = - 2.46837 \dots

Solution steps

3 x^{2} + 7 x = 1

Move

1

to the left side

3 x^{2} + 7 x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 3 ( - 1 )}{2 \cdot 3}

7^{2} - 4 \cdot 3 (- 1) = 61

x_{1, 2} = \frac{- 7 \pm 61}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 7 + 61}{2 \cdot 3}, x_{2} = \frac{- 7 - 61}{2 \cdot 3}

x = \frac{- 7 + 61}{2 \cdot 3} : \frac{- 7 + 61}{6}

x = \frac{- 7 - 61}{2 \cdot 3} : \frac{- 7 - 61}{6}

The solutions to the quadratic equation are:

x = \frac{- 7 + 61}{6}, x = \frac{- 7 - 61}{6}

x^{2} + 12 x + 35 = 0

x, k x^{2} + 4 y^{2} + 6 x - 8 y - 5 = 0

12 x^{2} + 11 x - 5 = 0

x, 3 x^{2} - 1 = 0

5 x^{2} + 5 x - 360 = 0

What are the solutions to the equation 3x^2+7x=1 ?
The solutions to the equation 3x^2+7x=1 are x=(-7+sqrt(61))/6 ,x=(-7-sqrt(61))/6
Find the zeros of 3x^2+7x=1
The zeros of 3x^2+7x=1 are x=(-7+sqrt(61))/6 ,x=(-7-sqrt(61))/6