What are the solutions to the equation 3x^2+12x+19=7 ?

The solutions to the equation 3x^2+12x+19=7 are x=-2

Find the zeros of 3x^2+12x+19=7

The zeros of 3x^2+12x+19=7 are x=-2

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3 x^{2} + 12 x + 19 = 7

x = - 2

Solution steps

3 x^{2} + 12 x + 19 = 7

Move

7

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

\frac{x}{100} = \frac{5089.3801}{9025.7957}

r^{2} + 4 r + 20 = 0

6 \leq - 4 x + 6 < 2

2 (x - 4) = 3 (1 + x)

81 z^{4} - 1

What are the solutions to the equation 3x^2+12x+19=7 ?
The solutions to the equation 3x^2+12x+19=7 are x=-2
Find the zeros of 3x^2+12x+19=7
The zeros of 3x^2+12x+19=7 are x=-2