What are the solutions to the equation 5c^2+7c+1=2c ?

The solutions to the equation 5c^2+7c+1=2c are c=(-sqrt(5)+1)/(2sqrt(5)),c=-(sqrt(5)+1)/(2sqrt(5))

Find the zeros of 5c^2+7c+1=2c

The zeros of 5c^2+7c+1=2c are c=(-sqrt(5)+1)/(2sqrt(5)),c=-(sqrt(5)+1)/(2sqrt(5))

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5 c^{2} + 7 c + 1 = 2 c

c = \frac{- 5 + 1}{2 5}, c = - \frac{5 + 1}{2 5}

Decimal Notation

c = - 0.27639 \dots, c = - 0.72360 \dots

Solution steps

5 c^{2} + 7 c + 1 = 2 c

Move

2 c

to the left side

5 c^{2} + 5 c + 1 = 0

Solve with the quadratic formula

c_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 5 \cdot 1}{2 \cdot 5}

5^{2} - 4 \cdot 5 \cdot 1 = 5

c_{1, 2} = \frac{- 5 \pm 5}{2 \cdot 5}

Separate the solutions

c_{1} = \frac{- 5 + 5}{2 \cdot 5}, c_{2} = \frac{- 5 - 5}{2 \cdot 5}

c = \frac{- 5 + 5}{2 \cdot 5} : \frac{- 5 + 1}{2 5}

c = \frac{- 5 - 5}{2 \cdot 5} : - \frac{5 + 1}{2 5}

The solutions to the quadratic equation are:

c = \frac{- 5 + 1}{2 5}, c = - \frac{5 + 1}{2 5}

q^{2} + 34 = 35

\frac{2}{10} - \frac{3}{10}

\frac{9 x ^{2} - 6 x + 1}{x ^{3} - x ^{2}} \leq 0

4 x^{2} + 10 x = - 3

2.61704 \cdot (\frac{1}{2})^{\frac{1000}{24100}}

What are the solutions to the equation 5c^2+7c+1=2c ?
The solutions to the equation 5c^2+7c+1=2c are c=(-sqrt(5)+1)/(2sqrt(5)),c=-(sqrt(5)+1)/(2sqrt(5))
Find the zeros of 5c^2+7c+1=2c
The zeros of 5c^2+7c+1=2c are c=(-sqrt(5)+1)/(2sqrt(5)),c=-(sqrt(5)+1)/(2sqrt(5))