What are the solutions to the equation 4c^2+c-6=0 ?

The solutions to the equation 4c^2+c-6=0 are c=(-1+sqrt(97))/8 ,c=(-1-sqrt(97))/8

Find the zeros of 4c^2+c-6=0

The zeros of 4c^2+c-6=0 are c=(-1+sqrt(97))/8 ,c=(-1-sqrt(97))/8

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4 c^{2} + c - 6 = 0

c = \frac{- 1 + 97}{8}, c = \frac{- 1 - 97}{8}

Decimal Notation

c = 1.10610 \dots, c = - 1.35610 \dots

Solution steps

4 c^{2} + c - 6 = 0

Solve with the quadratic formula

c_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 4 ( - 6 )}{2 \cdot 4}

1^{2} - 4 \cdot 4 (- 6) = 97

c_{1, 2} = \frac{- 1 \pm 97}{2 \cdot 4}

Separate the solutions

c_{1} = \frac{- 1 + 97}{2 \cdot 4}, c_{2} = \frac{- 1 - 97}{2 \cdot 4}

c = \frac{- 1 + 97}{2 \cdot 4} : \frac{- 1 + 97}{8}

c = \frac{- 1 - 97}{2 \cdot 4} : \frac{- 1 - 97}{8}

The solutions to the quadratic equation are:

c = \frac{- 1 + 97}{8}, c = \frac{- 1 - 97}{8}

9 x + 28 = 9 x^{2}

11 x = 3 x^{2} - 9

x, x^{2} - 4 x + 2^{t} = 0

x, x^{2} + y^{2} + 2 x = - 1

2 d^{2} = 8 d + 90

What are the solutions to the equation 4c^2+c-6=0 ?
The solutions to the equation 4c^2+c-6=0 are c=(-1+sqrt(97))/8 ,c=(-1-sqrt(97))/8
Find the zeros of 4c^2+c-6=0
The zeros of 4c^2+c-6=0 are c=(-1+sqrt(97))/8 ,c=(-1-sqrt(97))/8