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

The solutions to the equation 2c^2-4c=4 are c=1+sqrt(3),c=1-sqrt(3)

Find the zeros of 2c^2-4c=4

The zeros of 2c^2-4c=4 are c=1+sqrt(3),c=1-sqrt(3)

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2 c^{2} - 4 c = 4

c = 1 + 3, c = 1 - 3

Decimal Notation

c = 2.73205 \dots, c = - 0.73205 \dots

Solution steps

2 c^{2} - 4 c = 4

Move

4

to the left side

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

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 2 (- 4) = 43

c_{1, 2} = \frac{- ( - 4 ) \pm 4 3}{2 \cdot 2}

Separate the solutions

c_{1} = \frac{- ( - 4 ) + 4 3}{2 \cdot 2}, c_{2} = \frac{- ( - 4 ) - 4 3}{2 \cdot 2}

c = \frac{- ( - 4 ) + 4 3}{2 \cdot 2} : 1 + 3

c = \frac{- ( - 4 ) - 4 3}{2 \cdot 2} : 1 - 3

The solutions to the quadratic equation are:

c = 1 + 3, c = 1 - 3

x^{2} + 4 x = 12

x^{2} + 18 x - 111 = 0

3 x (4 x + 5) = 7 (4 x + 5)

w^{2} - 25 = 0

5 x^{2} - 15 x + 9 = 0

What are the solutions to the equation 2c^2-4c=4 ?
The solutions to the equation 2c^2-4c=4 are c=1+sqrt(3),c=1-sqrt(3)
Find the zeros of 2c^2-4c=4
The zeros of 2c^2-4c=4 are c=1+sqrt(3),c=1-sqrt(3)