What are the solutions to the equation 6x^2+12x+5=0 ?

The solutions to the equation 6x^2+12x+5=0 are x=(-6+sqrt(6))/6 ,x=-(sqrt(6)+1)/(sqrt(6))

Find the zeros of 6x^2+12x+5=0

The zeros of 6x^2+12x+5=0 are x=(-6+sqrt(6))/6 ,x=-(sqrt(6)+1)/(sqrt(6))

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6 x^{2} + 12 x + 5 = 0

x = \frac{- 6 + 6}{6}, x = - \frac{6 + 1}{6}

Solution steps

6 x^{2} + 12 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 6 \cdot 5}{2 \cdot 6}

1 2^{2} - 4 \cdot 6 \cdot 5 = 26

x_{1, 2} = \frac{- 12 \pm 2 6}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- 12 + 2 6}{2 \cdot 6}, x_{2} = \frac{- 12 - 2 6}{2 \cdot 6}

x = \frac{- 12 + 2 6}{2 \cdot 6} : \frac{- 6 + 6}{6}

x = \frac{- 12 - 2 6}{2 \cdot 6} : - \frac{6 + 1}{6}

The solutions to the quadratic equation are:

x = \frac{- 6 + 6}{6}, x = - \frac{6 + 1}{6}

9 x^{2} - 36 x - 45

\frac{y ^{7}}{y ^{3}}

\frac{9 x ^{3} y ^{3}}{36 x ^{5} y ^{6}}

\frac{7 π}{6} + 4 π

\frac{1}{2} \cdot \frac{4}{3}

What are the solutions to the equation 6x^2+12x+5=0 ?
The solutions to the equation 6x^2+12x+5=0 are x=(-6+sqrt(6))/6 ,x=-(sqrt(6)+1)/(sqrt(6))
Find the zeros of 6x^2+12x+5=0
The zeros of 6x^2+12x+5=0 are x=(-6+sqrt(6))/6 ,x=-(sqrt(6)+1)/(sqrt(6))