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

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

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

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

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

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

Decimal Notation

x = 0.86851 \dots, x = - 1.53518 \dots

Solution steps

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

Move

3 x

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}

\frac{12 a ^{3} + 33 a b ^{2} - 35 a ^{2} b - 10 ab}{4 a - 5 b}

4 - 3 (r + 2) = 6 - 8 (1 + 3 r)

b^{2} - 9 b + 8

lo g_{4} (256) = x

4 t^{3} + 3 t^{2} + 2 t

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