What are the solutions to the equation 2c^2-6c-2=3c ?

The solutions to the equation 2c^2-6c-2=3c are c=(9+sqrt(97))/4 ,c=(9-sqrt(97))/4

Find the zeros of 2c^2-6c-2=3c

The zeros of 2c^2-6c-2=3c are c=(9+sqrt(97))/4 ,c=(9-sqrt(97))/4

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2 c^{2} - 6 c - 2 = 3 c

c = \frac{9 + 97}{4}, c = \frac{9 - 97}{4}

Decimal Notation

c = 4.71221 \dots, c = - 0.21221 \dots

Solution steps

2 c^{2} - 6 c - 2 = 3 c

Move

3 c

to the left side

2 c^{2} - 9 c - 2 = 0

Solve with the quadratic formula

c_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 2 ( - 2 )}{2 \cdot 2}

(- 9)^{2} - 4 \cdot 2 (- 2) = 97

c_{1, 2} = \frac{- ( - 9 ) \pm 97}{2 \cdot 2}

Separate the solutions

c_{1} = \frac{- ( - 9 ) + 97}{2 \cdot 2}, c_{2} = \frac{- ( - 9 ) - 97}{2 \cdot 2}

c = \frac{- ( - 9 ) + 97}{2 \cdot 2} : \frac{9 + 97}{4}

c = \frac{- ( - 9 ) - 97}{2 \cdot 2} : \frac{9 - 97}{4}

The solutions to the quadratic equation are:

c = \frac{9 + 97}{4}, c = \frac{9 - 97}{4}

∣ 8 - 9 x ∣ = 8

\frac{- 3}{1 - x} < 7

3 + 5 x = 23

k^{2} - 8 k

- 23 \leq 4 x - 11 \leq 1

What are the solutions to the equation 2c^2-6c-2=3c ?
The solutions to the equation 2c^2-6c-2=3c are c=(9+sqrt(97))/4 ,c=(9-sqrt(97))/4
Find the zeros of 2c^2-6c-2=3c
The zeros of 2c^2-6c-2=3c are c=(9+sqrt(97))/4 ,c=(9-sqrt(97))/4