What are the solutions to the equation -3x^2+12x+1=0 ?

The solutions to the equation -3x^2+12x+1=0 are x=-(-6+sqrt(39))/3 ,x=(6+sqrt(39))/3

Find the zeros of -3x^2+12x+1=0

The zeros of -3x^2+12x+1=0 are x=-(-6+sqrt(39))/3 ,x=(6+sqrt(39))/3

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- 3 x^{2} + 12 x + 1 = 0

x = - \frac{- 6 + 39}{3}, x = \frac{6 + 39}{3}

Decimal Notation

x = - 0.08166 \dots, x = 4.08166 \dots

Solution steps

- 3 x^{2} + 12 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 ( - 3 ) \cdot 1}{2 ( - 3 )}

1 2^{2} - 4 (- 3) \cdot 1 = 239

x_{1, 2} = \frac{- 12 \pm 2 39}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- 12 + 2 39}{2 ( - 3 )}, x_{2} = \frac{- 12 - 2 39}{2 ( - 3 )}

x = \frac{- 12 + 2 39}{2 ( - 3 )} : - \frac{- 6 + 39}{3}

x = \frac{- 12 - 2 39}{2 ( - 3 )} : \frac{6 + 39}{3}

The solutions to the quadratic equation are:

x = - \frac{- 6 + 39}{3}, x = \frac{6 + 39}{3}

- 2.6 r = 2.08

\frac{( 6.5435 + ( 2 \cdot 6.5435 ) )}{2} (5)

\frac{4}{3} (x + \frac{1}{2}) = 10

\frac{π}{2}

4 x^{2} - 16 x + 14

What are the solutions to the equation -3x^2+12x+1=0 ?
The solutions to the equation -3x^2+12x+1=0 are x=-(-6+sqrt(39))/3 ,x=(6+sqrt(39))/3
Find the zeros of -3x^2+12x+1=0
The zeros of -3x^2+12x+1=0 are x=-(-6+sqrt(39))/3 ,x=(6+sqrt(39))/3