What are the solutions to the equation 3x^2+36x+120=12 ?

The solutions to the equation 3x^2+36x+120=12 are x=-6

Find the zeros of 3x^2+36x+120=12

The zeros of 3x^2+36x+120=12 are x=-6

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3 x^{2} + 36 x + 120 = 12

x = - 6

Solution steps

3 x^{2} + 36 x + 120 = 12

Move

12

to the left side

3 x^{2} + 36 x + 108 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 36 \pm 3 6 ^{2} - 4 \cdot 3 \cdot 108}{2 \cdot 3}

3 6^{2} - 4 \cdot 3 \cdot 108 = 0

x_{1, 2} = \frac{- 36 \pm 0}{2 \cdot 3}

x = \frac{- 36}{2 \cdot 3}

\frac{- 36}{2 \cdot 3} = - 6

x = - 6

The solution to the quadratic equation is:

x = - 6

\frac{1}{9} y^{2} = y + \frac{1}{36}

2 x^{2} - 8 x = 12

4 x = 60

- 161

0.2 5^{- 2}

What are the solutions to the equation 3x^2+36x+120=12 ?
The solutions to the equation 3x^2+36x+120=12 are x=-6
Find the zeros of 3x^2+36x+120=12
The zeros of 3x^2+36x+120=12 are x=-6