What are the solutions to the equation 12x^2+18x+6=0 ?

The solutions to the equation 12x^2+18x+6=0 are x=-1/2 ,x=-1

Find the zeros of 12x^2+18x+6=0

The zeros of 12x^2+18x+6=0 are x=-1/2 ,x=-1

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12 x^{2} + 18 x + 6 = 0

x = - \frac{1}{2}, x = - 1

Solution steps

12 x^{2} + 18 x + 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 \cdot 12 \cdot 6}{2 \cdot 12}

1 8^{2} - 4 \cdot 12 \cdot 6 = 6

x_{1, 2} = \frac{- 18 \pm 6}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- 18 + 6}{2 \cdot 12}, x_{2} = \frac{- 18 - 6}{2 \cdot 12}

x = \frac{- 18 + 6}{2 \cdot 12} : - \frac{1}{2}

x = \frac{- 18 - 6}{2 \cdot 12} : - 1

The solutions to the quadratic equation are:

x = - \frac{1}{2}, x = - 1

(8 x - 10) (x - 1) = 0

x^{2} + 56 x = 8.16

x, 2 x^{2} + 7 x + 5 = 0

x (4 x - 1) = 5

u, A = 28 u^{2}

What are the solutions to the equation 12x^2+18x+6=0 ?
The solutions to the equation 12x^2+18x+6=0 are x=-1/2 ,x=-1
Find the zeros of 12x^2+18x+6=0
The zeros of 12x^2+18x+6=0 are x=-1/2 ,x=-1