What are the solutions to the equation -3x^2+2x+4=0 ?

The solutions to the equation -3x^2+2x+4=0 are x=-(-1+sqrt(13))/3 ,x=(1+sqrt(13))/3

Find the zeros of -3x^2+2x+4=0

The zeros of -3x^2+2x+4=0 are x=-(-1+sqrt(13))/3 ,x=(1+sqrt(13))/3

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

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} + 2 x + 4 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

(\frac{1}{32})^{\frac{2}{5}}

\frac{5 - 3 x}{2} - (\frac{1}{4} x - \frac{4 x - 3}{2}) = - 2

x - 3 = 7

q, q + r + s = P

a^{2} - 4 a - 24 + 6 a

What are the solutions to the equation -3x^2+2x+4=0 ?
The solutions to the equation -3x^2+2x+4=0 are x=-(-1+sqrt(13))/3 ,x=(1+sqrt(13))/3
Find the zeros of -3x^2+2x+4=0
The zeros of -3x^2+2x+4=0 are x=-(-1+sqrt(13))/3 ,x=(1+sqrt(13))/3