What are the solutions to the equation 3x^2+36x+81=0 ?

The solutions to the equation 3x^2+36x+81=0 are x=-3,x=-9

Find the zeros of 3x^2+36x+81=0

The zeros of 3x^2+36x+81=0 are x=-3,x=-9

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3 x^{2} + 36 x + 81 = 0

x = - 3, x = - 9

Solution steps

3 x^{2} + 36 x + 81 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 36 \pm 3 6 ^{2} - 4 \cdot 3 \cdot 81}{2 \cdot 3}

3 6^{2} - 4 \cdot 3 \cdot 81 = 18

x_{1, 2} = \frac{- 36 \pm 18}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 36 + 18}{2 \cdot 3}, x_{2} = \frac{- 36 - 18}{2 \cdot 3}

x = \frac{- 36 + 18}{2 \cdot 3} : - 3

x = \frac{- 36 - 18}{2 \cdot 3} : - 9

The solutions to the quadratic equation are:

x = - 3, x = - 9

∣ 2 (x - 1) + 4 ∣ \leq 26

x^{2} = x + 132

5 x + \frac{5}{3} \leq - 2 x - \frac{2}{3}

5 x + 2 \leq 27

- 35 + 6 x = 7 (x - 4)

What are the solutions to the equation 3x^2+36x+81=0 ?
The solutions to the equation 3x^2+36x+81=0 are x=-3,x=-9
Find the zeros of 3x^2+36x+81=0
The zeros of 3x^2+36x+81=0 are x=-3,x=-9