What are the solutions to the equation solvefor x,4x^2+12x+3=0 ?

The solutions to the equation solvefor x,4x^2+12x+3=0 are x=(-3+sqrt(6))/2 ,x=-(3+sqrt(6))/2

Find the zeros of solvefor x,4x^2+12x+3=0

The zeros of solvefor x,4x^2+12x+3=0 are x=(-3+sqrt(6))/2 ,x=-(3+sqrt(6))/2

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x, 4 x^{2} + 12 x + 3 = 0

x = \frac{- 3 + 6}{2}, x = - \frac{3 + 6}{2}

Decimal Notation

x = - 0.27525 \dots, x = - 2.72474 \dots

Solution steps

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

Solve with the quadratic formula

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

1 2^{2} - 4 \cdot 4 \cdot 3 = 46

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

Separate the solutions

x_{1} = \frac{- 12 + 4 6}{2 \cdot 4}, x_{2} = \frac{- 12 - 4 6}{2 \cdot 4}

x = \frac{- 12 + 4 6}{2 \cdot 4} : \frac{- 3 + 6}{2}

x = \frac{- 12 - 4 6}{2 \cdot 4} : - \frac{3 + 6}{2}

The solutions to the quadratic equation are:

x = \frac{- 3 + 6}{2}, x = - \frac{3 + 6}{2}

2 x^{2} = 7 x + 15

9 = π + r^{2} 16

2 x^{2} = x + 8

(x + 4)^{2} = 31

(5 p + 2)^{2} = 8

What are the solutions to the equation solvefor x,4x^2+12x+3=0 ?
The solutions to the equation solvefor x,4x^2+12x+3=0 are x=(-3+sqrt(6))/2 ,x=-(3+sqrt(6))/2
Find the zeros of solvefor x,4x^2+12x+3=0
The zeros of solvefor x,4x^2+12x+3=0 are x=(-3+sqrt(6))/2 ,x=-(3+sqrt(6))/2