What are the solutions to the equation 3x^2+2x=10 ?

The solutions to the equation 3x^2+2x=10 are x=(-1+sqrt(31))/3 ,x=-(1+sqrt(31))/3

Find the zeros of 3x^2+2x=10

The zeros of 3x^2+2x=10 are x=(-1+sqrt(31))/3 ,x=-(1+sqrt(31))/3

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3 x^{2} + 2 x = 10

x = \frac{- 1 + 31}{3}, x = - \frac{1 + 31}{3}

Decimal Notation

x = 1.52258 \dots, x = - 2.18925 \dots

Solution steps

3 x^{2} + 2 x = 10

Move

10

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 3 ( - 10 )}{2 \cdot 3}

2^{2} - 4 \cdot 3 (- 10) = 231

x_{1, 2} = \frac{- 2 \pm 2 31}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 2 + 2 31}{2 \cdot 3}, x_{2} = \frac{- 2 - 2 31}{2 \cdot 3}

x = \frac{- 2 + 2 31}{2 \cdot 3} : \frac{- 1 + 31}{3}

x = \frac{- 2 - 2 31}{2 \cdot 3} : - \frac{1 + 31}{3}

The solutions to the quadratic equation are:

x = \frac{- 1 + 31}{3}, x = - \frac{1 + 31}{3}

x^{\frac{5}{2}} - 100 x^{\frac{1}{2}}

b + (2 b + 13) = 76

2 x (x - 4) + 5 x (1 - x) = 3 (x + 1)

∣ x - 4 ∣ \leq 7

4 x + 1 = x + 7

What are the solutions to the equation 3x^2+2x=10 ?
The solutions to the equation 3x^2+2x=10 are x=(-1+sqrt(31))/3 ,x=-(1+sqrt(31))/3
Find the zeros of 3x^2+2x=10
The zeros of 3x^2+2x=10 are x=(-1+sqrt(31))/3 ,x=-(1+sqrt(31))/3