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

The solutions to the equation 2x^2+12x+3=0 are x=(-6+sqrt(30))/2 ,x=-(6+sqrt(30))/2

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

The zeros of 2x^2+12x+3=0 are x=(-6+sqrt(30))/2 ,x=-(6+sqrt(30))/2

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

x = \frac{- 6 + 30}{2}, x = - \frac{6 + 30}{2}

Decimal Notation

x = - 0.26138 \dots, x = - 5.73861 \dots

Solution steps

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

Solve with the quadratic formula

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

1 2^{2} - 4 \cdot 2 \cdot 3 = 230

x_{1, 2} = \frac{- 12 \pm 2 30}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 12 + 2 30}{2 \cdot 2}, x_{2} = \frac{- 12 - 2 30}{2 \cdot 2}

x = \frac{- 12 + 2 30}{2 \cdot 2} : \frac{- 6 + 30}{2}

x = \frac{- 12 - 2 30}{2 \cdot 2} : - \frac{6 + 30}{2}

The solutions to the quadratic equation are:

x = \frac{- 6 + 30}{2}, x = - \frac{6 + 30}{2}

2 x^{3} + 3 x^{2} - 4 x - 6

\frac{r}{17} = 8

\frac{∣ x - 3 ∣}{∣ 7 - x ∣} > 0

3 (x + 2) + 4 (x - 5) = 10

\frac{15 x ^{5} y ^{5}}{3 x y ^{8}}

What are the solutions to the equation 2x^2+12x+3=0 ?
The solutions to the equation 2x^2+12x+3=0 are x=(-6+sqrt(30))/2 ,x=-(6+sqrt(30))/2
Find the zeros of 2x^2+12x+3=0
The zeros of 2x^2+12x+3=0 are x=(-6+sqrt(30))/2 ,x=-(6+sqrt(30))/2