What are the solutions to the equation 3c^2+7c+2=2c^2 ?

The solutions to the equation 3c^2+7c+2=2c^2 are c=(-7+sqrt(41))/2 ,c=(-7-sqrt(41))/2

Find the zeros of 3c^2+7c+2=2c^2

The zeros of 3c^2+7c+2=2c^2 are c=(-7+sqrt(41))/2 ,c=(-7-sqrt(41))/2

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3 c^{2} + 7 c + 2 = 2 c^{2}

c = \frac{- 7 + 41}{2}, c = \frac{- 7 - 41}{2}

Decimal Notation

c = - 0.29843 \dots, c = - 6.70156 \dots

Solution steps

3 c^{2} + 7 c + 2 = 2 c^{2}

Move

2 c^{2}

to the left side

c^{2} + 7 c + 2 = 0

Solve with the quadratic formula

c_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot 2}{2 \cdot 1}

7^{2} - 4 \cdot 1 \cdot 2 = 41

c_{1, 2} = \frac{- 7 \pm 41}{2 \cdot 1}

Separate the solutions

c_{1} = \frac{- 7 + 41}{2 \cdot 1}, c_{2} = \frac{- 7 - 41}{2 \cdot 1}

c = \frac{- 7 + 41}{2 \cdot 1} : \frac{- 7 + 41}{2}

c = \frac{- 7 - 41}{2 \cdot 1} : \frac{- 7 - 41}{2}

The solutions to the quadratic equation are:

c = \frac{- 7 + 41}{2}, c = \frac{- 7 - 41}{2}

x^{2} + 10 x - 25 = 4 x + 2

s^{2} + s + 1

5 x^{2} - 36 x + 36 = 0

(x + 3) (x - 6) = 0

x^{2} + 3 x - 5

What are the solutions to the equation 3c^2+7c+2=2c^2 ?
The solutions to the equation 3c^2+7c+2=2c^2 are c=(-7+sqrt(41))/2 ,c=(-7-sqrt(41))/2
Find the zeros of 3c^2+7c+2=2c^2
The zeros of 3c^2+7c+2=2c^2 are c=(-7+sqrt(41))/2 ,c=(-7-sqrt(41))/2