What are the solutions to the equation 3x^2+2x=4 ?

The solutions to the equation 3x^2+2x=4 are x=(-1+sqrt(13))/3 ,x=-(1+sqrt(13))/3

Find the zeros of 3x^2+2x=4

The zeros of 3x^2+2x=4 are x=(-1+sqrt(13))/3 ,x=-(1+sqrt(13))/3

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3 x^{2} + 2 x = 4

x = \frac{- 1 + 13}{3}, x = - \frac{1 + 13}{3}

Solution steps

3 x^{2} + 2 x = 4

Move

4

to the left side

3 x^{2} + 2 x - 4 = 0

Solve with the quadratic formula

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

2^{2} - 4 \cdot 3 (- 4) = 213

x_{1, 2} = \frac{- 2 \pm 2 13}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 2 + 2 13}{2 \cdot 3}, x_{2} = \frac{- 2 - 2 13}{2 \cdot 3}

x = \frac{- 2 + 2 13}{2 \cdot 3} : \frac{- 1 + 13}{3}

x = \frac{- 2 - 2 13}{2 \cdot 3} : - \frac{1 + 13}{3}

The solutions to the quadratic equation are:

x = \frac{- 1 + 13}{3}, x = - \frac{1 + 13}{3}

1 + 12 x + 36 x^{2}

5 x = 3

(x + 1)^{2} = - 3

\frac{( 2 x - 3 ) ( 3 x + 8 )}{( x - 8 ) ^{3}} \geq 0

- 4 \leq 5 x + 13 < 23

What are the solutions to the equation 3x^2+2x=4 ?
The solutions to the equation 3x^2+2x=4 are x=(-1+sqrt(13))/3 ,x=-(1+sqrt(13))/3
Find the zeros of 3x^2+2x=4
The zeros of 3x^2+2x=4 are x=(-1+sqrt(13))/3 ,x=-(1+sqrt(13))/3