What are the solutions to the equation 27c^2+108c+108=0 ?

The solutions to the equation 27c^2+108c+108=0 are c=-2

Find the zeros of 27c^2+108c+108=0

The zeros of 27c^2+108c+108=0 are c=-2

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27 c^{2} + 108 c + 108 = 0

c = - 2

Solution steps

27 c^{2} + 108 c + 108 = 0

Solve with the quadratic formula

c_{1, 2} = \frac{- 108 \pm 10 8 ^{2} - 4 \cdot 27 \cdot 108}{2 \cdot 27}

10 8^{2} - 4 \cdot 27 \cdot 108 = 0

c_{1, 2} = \frac{- 108 \pm 0}{2 \cdot 27}

c = \frac{- 108}{2 \cdot 27}

\frac{- 108}{2 \cdot 27} = - 2

c = - 2

The solution to the quadratic equation is:

c = - 2

\frac{24 !}{4 ! ( 24 - 4 ) !}

\frac{- 5 + x}{- 3} = - 5

x^{2} = (x + 3) (x - 3) + 9

\frac{2 x ^{2} + 8 y}{6 y ^{3} + 4 x ^{2}}

x^{2} + 6 x - 13 = 0

What are the solutions to the equation 27c^2+108c+108=0 ?
The solutions to the equation 27c^2+108c+108=0 are c=-2
Find the zeros of 27c^2+108c+108=0
The zeros of 27c^2+108c+108=0 are c=-2