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

The solutions to the equation 2c^2+12c+11=0 are c=(-6+sqrt(14))/2 ,c=-(6+sqrt(14))/2

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

The zeros of 2c^2+12c+11=0 are c=(-6+sqrt(14))/2 ,c=-(6+sqrt(14))/2

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

c = \frac{- 6 + 14}{2}, c = - \frac{6 + 14}{2}

Decimal

c = - 1.12917 \dots, c = - 4.87082 \dots

Solution steps

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

Solve with the quadratic formula

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

1 2^{2} - 4 \cdot 2 \cdot 11 = 214

c_{1, 2} = \frac{- 12 \pm 2 14}{2 \cdot 2}

Separate the solutions

c_{1} = \frac{- 12 + 2 14}{2 \cdot 2}, c_{2} = \frac{- 12 - 2 14}{2 \cdot 2}

c = \frac{- 12 + 2 14}{2 \cdot 2} : \frac{- 6 + 14}{2}

c = \frac{- 12 - 2 14}{2 \cdot 2} : - \frac{6 + 14}{2}

The solutions to the quadratic equation are:

c = \frac{- 6 + 14}{2}, c = - \frac{6 + 14}{2}

5 (3 x + 6)^{2} + 9 = 44

x^{2} - 12 x + 52

2 x^{2} + 1

7 x^{2} + 36 x - 1440 = 0

y, x^{2} + 14 x y + 49 y^{2} = 100

What are the solutions to the equation 2c^2+12c+11=0 ?
The solutions to the equation 2c^2+12c+11=0 are c=(-6+sqrt(14))/2 ,c=-(6+sqrt(14))/2
Find the zeros of 2c^2+12c+11=0
The zeros of 2c^2+12c+11=0 are c=(-6+sqrt(14))/2 ,c=-(6+sqrt(14))/2