What are the solutions to the equation 3x^2+12x+18=6 ?

The solutions to the equation 3x^2+12x+18=6 are x=-2

Find the zeros of 3x^2+12x+18=6

The zeros of 3x^2+12x+18=6 are x=-2

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

x = - 2

Solution steps

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

Move

6

to the left side

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

Solve with the quadratic formula

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

1 2^{2} - 4 \cdot 3 \cdot 12 = 0

x_{1, 2} = \frac{- 12 \pm 0}{2 \cdot 3}

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

\frac{- 12}{2 \cdot 3} = - 2

x = - 2

The solution to the quadratic equation is:

x = - 2

3 y^{2} - 6 y - 24

\frac{4 ^{n + 1}}{( 2 ^{n - 1} ) ^{n + 1}}

\frac{x + 4}{2 x - 1} \leq 3

\frac{x + 2}{x - 4} < 1

- x^{2} + 4 x + 3 = ∣ x ∣ - 2

What are the solutions to the equation 3x^2+12x+18=6 ?
The solutions to the equation 3x^2+12x+18=6 are x=-2
Find the zeros of 3x^2+12x+18=6
The zeros of 3x^2+12x+18=6 are x=-2