What are the solutions to the equation 27c^2-54c+18=0 ?

The solutions to the equation 27c^2-54c+18=0 are c=(3+sqrt(3))/3 ,c=(3-sqrt(3))/3

Find the zeros of 27c^2-54c+18=0

The zeros of 27c^2-54c+18=0 are c=(3+sqrt(3))/3 ,c=(3-sqrt(3))/3

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27 c^{2} - 54 c + 18 = 0

c = \frac{3 + 3}{3}, c = \frac{3 - 3}{3}

Decimal Notation

c = 1.57735 \dots, c = 0.42264 \dots

Solution steps

27 c^{2} - 54 c + 18 = 0

Solve with the quadratic formula

c_{1, 2} = \frac{- ( - 54 ) \pm ( - 54 ) ^{2} - 4 \cdot 27 \cdot 18}{2 \cdot 27}

(- 54)^{2} - 4 \cdot 27 \cdot 18 = 183

c_{1, 2} = \frac{- ( - 54 ) \pm 18 3}{2 \cdot 27}

Separate the solutions

c_{1} = \frac{- ( - 54 ) + 18 3}{2 \cdot 27}, c_{2} = \frac{- ( - 54 ) - 18 3}{2 \cdot 27}

c = \frac{- ( - 54 ) + 18 3}{2 \cdot 27} : \frac{3 + 3}{3}

c = \frac{- ( - 54 ) - 18 3}{2 \cdot 27} : \frac{3 - 3}{3}

The solutions to the quadratic equation are:

c = \frac{3 + 3}{3}, c = \frac{3 - 3}{3}

5 x^{2} - 5 x - 4 = 0

5 x^{2} - 18 = 0

- 3 x^{2} + 24 x = 15

\frac{5}{4} q^{2} = - \frac{5}{4}

100 r^{2} - 5 = 76

What are the solutions to the equation 27c^2-54c+18=0 ?
The solutions to the equation 27c^2-54c+18=0 are c=(3+sqrt(3))/3 ,c=(3-sqrt(3))/3
Find the zeros of 27c^2-54c+18=0
The zeros of 27c^2-54c+18=0 are c=(3+sqrt(3))/3 ,c=(3-sqrt(3))/3