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(66))/6 ,x=-(6+sqrt(66))/6

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

The zeros of 6x^2+12x-5=0 are x=(-6+sqrt(66))/6 ,x=-(6+sqrt(66))/6

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

x = \frac{- 6 + 66}{6}, x = - \frac{6 + 66}{6}

Decimal Notation

x = 0.35400 \dots, x = - 2.35400 \dots

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 ( - 5 )}{2 \cdot 6}

1 2^{2} - 4 \cdot 6 (- 5) = 266

x_{1, 2} = \frac{- 12 \pm 2 66}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- 12 + 2 66}{2 \cdot 6}, x_{2} = \frac{- 12 - 2 66}{2 \cdot 6}

x = \frac{- 12 + 2 66}{2 \cdot 6} : \frac{- 6 + 66}{6}

x = \frac{- 12 - 2 66}{2 \cdot 6} : - \frac{6 + 66}{6}

The solutions to the quadratic equation are:

x = \frac{- 6 + 66}{6}, x = - \frac{6 + 66}{6}

\frac{5 x ^{8} - 20 x ^{4}}{5 x ^{8} + 20 x ^{6} + 20 x ^{4}}

8 < 2 (a + 3) < 10

x^{2} + 3 x + 5 < 5

\frac{1}{1 - x} \leq \frac{3}{x}

(6^{5})^{4}

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(66))/6 ,x=-(6+sqrt(66))/6
Find the zeros of 6x^2+12x-5=0
The zeros of 6x^2+12x-5=0 are x=(-6+sqrt(66))/6 ,x=-(6+sqrt(66))/6