What are the solutions to the equation 3c^2+12c+11=0 ?

The solutions to the equation 3c^2+12c+11=0 are c=(-6+sqrt(3))/3 ,c=-(6+sqrt(3))/3

Find the zeros of 3c^2+12c+11=0

The zeros of 3c^2+12c+11=0 are c=(-6+sqrt(3))/3 ,c=-(6+sqrt(3))/3

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3 c^{2} + 12 c + 11 = 0

c = \frac{- 6 + 3}{3}, c = - \frac{6 + 3}{3}

Decimal Notation

c = - 1.42264 \dots, c = - 2.57735 \dots

Solution steps

3 c^{2} + 12 c + 11 = 0

Solve with the quadratic formula

c_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 3 \cdot 11}{2 \cdot 3}

1 2^{2} - 4 \cdot 3 \cdot 11 = 23

c_{1, 2} = \frac{- 12 \pm 2 3}{2 \cdot 3}

Separate the solutions

c_{1} = \frac{- 12 + 2 3}{2 \cdot 3}, c_{2} = \frac{- 12 - 2 3}{2 \cdot 3}

c = \frac{- 12 + 2 3}{2 \cdot 3} : \frac{- 6 + 3}{3}

c = \frac{- 12 - 2 3}{2 \cdot 3} : - \frac{6 + 3}{3}

The solutions to the quadratic equation are:

c = \frac{- 6 + 3}{3}, c = - \frac{6 + 3}{3}

2 y^{2} - 6 y + 8 = 0

x^{2} = 40 - 3 x

4 x^{2} - 40 x = - 24

3 x^{2} - 17 x + 24 = 0

x^{2} - 11 x + 8 = 0

What are the solutions to the equation 3c^2+12c+11=0 ?
The solutions to the equation 3c^2+12c+11=0 are c=(-6+sqrt(3))/3 ,c=-(6+sqrt(3))/3
Find the zeros of 3c^2+12c+11=0
The zeros of 3c^2+12c+11=0 are c=(-6+sqrt(3))/3 ,c=-(6+sqrt(3))/3