What are the solutions to the equation -5c^2+c-18=-6c^2 ?

The solutions to the equation -5c^2+c-18=-6c^2 are c=(-1+sqrt(73))/2 ,c=(-1-sqrt(73))/2

Find the zeros of -5c^2+c-18=-6c^2

The zeros of -5c^2+c-18=-6c^2 are c=(-1+sqrt(73))/2 ,c=(-1-sqrt(73))/2

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- 5 c^{2} + c - 18 = - 6 c^{2}

c = \frac{- 1 + 73}{2}, c = \frac{- 1 - 73}{2}

Decimal Notation

c = 3.77200 \dots, c = - 4.77200 \dots

Solution steps

- 5 c^{2} + c - 18 = - 6 c^{2}

Move

6 c^{2}

to the left side

c^{2} + c - 18 = 0

Solve with the quadratic formula

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

1^{2} - 4 \cdot 1 \cdot (- 18) = 73

c_{1, 2} = \frac{- 1 \pm 73}{2 \cdot 1}

Separate the solutions

c_{1} = \frac{- 1 + 73}{2 \cdot 1}, c_{2} = \frac{- 1 - 73}{2 \cdot 1}

c = \frac{- 1 + 73}{2 \cdot 1} : \frac{- 1 + 73}{2}

c = \frac{- 1 - 73}{2 \cdot 1} : \frac{- 1 - 73}{2}

The solutions to the quadratic equation are:

c = \frac{- 1 + 73}{2}, c = \frac{- 1 - 73}{2}

(2 x - 1) (x + 2) = (x + 3) (x + 1)

y^{2} - 2 y - 1 = 0

5 x^{2} = 7 x + 7

16 (5 x - 3)^{2} - 28 = - 19

(x + 12)^{2} = 9

What are the solutions to the equation -5c^2+c-18=-6c^2 ?
The solutions to the equation -5c^2+c-18=-6c^2 are c=(-1+sqrt(73))/2 ,c=(-1-sqrt(73))/2
Find the zeros of -5c^2+c-18=-6c^2
The zeros of -5c^2+c-18=-6c^2 are c=(-1+sqrt(73))/2 ,c=(-1-sqrt(73))/2