What are the solutions to the equation 0=3x^2+12x+9 ?

The solutions to the equation 0=3x^2+12x+9 are x=-1,x=-3

Find the zeros of 0=3x^2+12x+9

The zeros of 0=3x^2+12x+9 are x=-1,x=-3

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0 = 3 x^{2} + 12 x + 9

x = - 1, x = - 3

Solution steps

0 = 3 x^{2} + 12 x + 9

Switch sides

3 x^{2} + 12 x + 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 3 \cdot 9}{2 \cdot 3}

1 2^{2} - 4 \cdot 3 \cdot 9 = 6

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

Separate the solutions

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

x = \frac{- 12 + 6}{2 \cdot 3} : - 1

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

The solutions to the quadratic equation are:

x = - 1, x = - 3

z, z^{2} = x^{2} + y^{2}

2 n^{2} - 52 = n^{2} - 3

6 x^{2} = x - 1

t^{2} = 45

t^{2} = 36

What are the solutions to the equation 0=3x^2+12x+9 ?
The solutions to the equation 0=3x^2+12x+9 are x=-1,x=-3
Find the zeros of 0=3x^2+12x+9
The zeros of 0=3x^2+12x+9 are x=-1,x=-3