What are the solutions to the equation 3x^2+12x+17=5 ?

The solutions to the equation 3x^2+12x+17=5 are x=-2

Find the zeros of 3x^2+12x+17=5

The zeros of 3x^2+12x+17=5 are x=-2

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3 x^{2} + 12 x + 17 = 5

x = - 2

Solution steps

3 x^{2} + 12 x + 17 = 5

Move

5

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{1}{x} + \frac{1}{1 - x} > 0

6 x^{2} + 12 x

- ∣ 4 x - 7 ∣ \geq 3

(2 a + 3)^{2} - (a - 1)^{2}

25 6^{\frac{3}{2}}

What are the solutions to the equation 3x^2+12x+17=5 ?
The solutions to the equation 3x^2+12x+17=5 are x=-2
Find the zeros of 3x^2+12x+17=5
The zeros of 3x^2+12x+17=5 are x=-2