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

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

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

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

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

x = - 2 + 3, x = - 2 - 3

Decimal Notation

x = - 0.26794 \dots, x = - 3.73205 \dots

Solution steps

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

Solve with the quadratic formula

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

1 2^{2} - 4 \cdot 3 \cdot 3 = 63

x_{1, 2} = \frac{- 12 \pm 6 3}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 12 + 6 3}{2 \cdot 3}, x_{2} = \frac{- 12 - 6 3}{2 \cdot 3}

x = \frac{- 12 + 6 3}{2 \cdot 3} : - 2 + 3

x = \frac{- 12 - 6 3}{2 \cdot 3} : - 2 - 3

The solutions to the quadratic equation are:

x = - 2 + 3, x = - 2 - 3

2 (- 3) + \frac{3}{( - 2 )} + \frac{- ( 2 )}{( - 16 )}

\frac{x}{5} - \frac{x}{9} = - 1

3 y + 1 < 3 y + 1

(x^{\frac{1}{2}})^{\frac{3}{2}}

(\frac{2 x}{y ^{4}})^{2}

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