What are the solutions to the equation -6g^2+3g+5=0 ?

The solutions to the equation -6g^2+3g+5=0 are g=-(-3+sqrt(129))/(12),g=(3+sqrt(129))/(12)

Find the zeros of -6g^2+3g+5=0

The zeros of -6g^2+3g+5=0 are g=-(-3+sqrt(129))/(12),g=(3+sqrt(129))/(12)

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- 6 g^{2} + 3 g + 5 = 0

g = - \frac{- 3 + 129}{12}, g = \frac{3 + 129}{12}

Solution steps

- 6 g^{2} + 3 g + 5 = 0

Solve with the quadratic formula

g_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 6 ) \cdot 5}{2 ( - 6 )}

3^{2} - 4 (- 6) \cdot 5 = 129

g_{1, 2} = \frac{- 3 \pm 129}{2 ( - 6 )}

Separate the solutions

g_{1} = \frac{- 3 + 129}{2 ( - 6 )}, g_{2} = \frac{- 3 - 129}{2 ( - 6 )}

g = \frac{- 3 + 129}{2 ( - 6 )} : - \frac{- 3 + 129}{12}

g = \frac{- 3 - 129}{2 ( - 6 )} : \frac{3 + 129}{12}

The solutions to the quadratic equation are:

g = - \frac{- 3 + 129}{12}, g = \frac{3 + 129}{12}

1 + \frac{10 x}{11} - \frac{1}{10} (x + 11) \geq x

\frac{\frac{1}{x} + \frac{1}{y}}{\frac{1}{x} - \frac{1}{y}}

6 (v - 5) - 2 v = 2

(\frac{1}{2}) (\frac{1}{2})

1 0^{5 l o g_{10} (5)}

What are the solutions to the equation -6g^2+3g+5=0 ?
The solutions to the equation -6g^2+3g+5=0 are g=-(-3+sqrt(129))/(12),g=(3+sqrt(129))/(12)
Find the zeros of -6g^2+3g+5=0
The zeros of -6g^2+3g+5=0 are g=-(-3+sqrt(129))/(12),g=(3+sqrt(129))/(12)