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

The solutions to the equation 3x^2+7x=2 are x=(-7+sqrt(73))/6 ,x=(-7-sqrt(73))/6

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

The zeros of 3x^2+7x=2 are x=(-7+sqrt(73))/6 ,x=(-7-sqrt(73))/6

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3 x^{2} + 7 x = 2

x = \frac{- 7 + 73}{6}, x = \frac{- 7 - 73}{6}

Solution steps

3 x^{2} + 7 x = 2

Move

2

to the left side

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

Solve with the quadratic formula

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

7^{2} - 4 \cdot 3 (- 2) = 73

x_{1, 2} = \frac{- 7 \pm 73}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 7 + 73}{2 \cdot 3}, x_{2} = \frac{- 7 - 73}{2 \cdot 3}

x = \frac{- 7 + 73}{2 \cdot 3} : \frac{- 7 + 73}{6}

x = \frac{- 7 - 73}{2 \cdot 3} : \frac{- 7 - 73}{6}

The solutions to the quadratic equation are:

x = \frac{- 7 + 73}{6}, x = \frac{- 7 - 73}{6}

409 6^{\frac{1}{6}}

- 3 n (2 n - 7) = 0

∣ x - 3 ∣ \geq 5

25 x^{2} y^{2} - 4 x^{2} y + 16 x^{2}

\frac{1}{x - 2 - 4} \leq 0

What are the solutions to the equation 3x^2+7x=2 ?
The solutions to the equation 3x^2+7x=2 are x=(-7+sqrt(73))/6 ,x=(-7-sqrt(73))/6
Find the zeros of 3x^2+7x=2
The zeros of 3x^2+7x=2 are x=(-7+sqrt(73))/6 ,x=(-7-sqrt(73))/6