What are the solutions to the equation 2c^2-12c+7=-4c ?

The solutions to the equation 2c^2-12c+7=-4c are c=(4+sqrt(2))/2 ,c=(4-sqrt(2))/2

Find the zeros of 2c^2-12c+7=-4c

The zeros of 2c^2-12c+7=-4c are c=(4+sqrt(2))/2 ,c=(4-sqrt(2))/2

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2 c^{2} - 12 c + 7 = - 4 c

c = \frac{4 + 2}{2}, c = \frac{4 - 2}{2}

Decimal Notation

c = 2.70710 \dots, c = 1.29289 \dots

Solution steps

2 c^{2} - 12 c + 7 = - 4 c

Move

4 c

to the left side

2 c^{2} - 8 c + 7 = 0

Solve with the quadratic formula

c_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 2 \cdot 7}{2 \cdot 2}

(- 8)^{2} - 4 \cdot 2 \cdot 7 = 22

c_{1, 2} = \frac{- ( - 8 ) \pm 2 2}{2 \cdot 2}

Separate the solutions

c_{1} = \frac{- ( - 8 ) + 2 2}{2 \cdot 2}, c_{2} = \frac{- ( - 8 ) - 2 2}{2 \cdot 2}

c = \frac{- ( - 8 ) + 2 2}{2 \cdot 2} : \frac{4 + 2}{2}

c = \frac{- ( - 8 ) - 2 2}{2 \cdot 2} : \frac{4 - 2}{2}

The solutions to the quadratic equation are:

c = \frac{4 + 2}{2}, c = \frac{4 - 2}{2}

x^{2} - 7

4 y^{2} + 3 y - 1 = 0

x, 5 a x^{2} y - 5 x^{2} + 6 y^{3} = 100

(3 x + 1) (3 x - 1) = 5 x + 3

2 a^{2} - 5 a - 2 = - 3 a^{2}

What are the solutions to the equation 2c^2-12c+7=-4c ?
The solutions to the equation 2c^2-12c+7=-4c are c=(4+sqrt(2))/2 ,c=(4-sqrt(2))/2
Find the zeros of 2c^2-12c+7=-4c
The zeros of 2c^2-12c+7=-4c are c=(4+sqrt(2))/2 ,c=(4-sqrt(2))/2