What are the solutions to the equation 2x^2-4x+c=0 ?

The solutions to the equation 2x^2-4x+c=0 are x=(4+sqrt(16-8c))/4 ,x=(4-sqrt(16-8c))/4

Find the zeros of 2x^2-4x+c=0

The zeros of 2x^2-4x+c=0 are x=(4+sqrt(16-8c))/4 ,x=(4-sqrt(16-8c))/4

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

x = \frac{4 + 16 - 8 c}{4}, x = \frac{4 - 16 - 8 c}{4}

Solution steps

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

Solve with the quadratic formula

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

Simplify

(- 4)^{2} - 4 \cdot 2 c : 16 - 8 c

x_{1, 2} = \frac{- ( - 4 ) \pm 16 - 8 c}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 16 - 8 c}{2 \cdot 2}, x_{2} = \frac{- ( - 4 ) - 16 - 8 c}{2 \cdot 2}

x = \frac{- ( - 4 ) + 16 - 8 c}{2 \cdot 2} : \frac{4 + 16 - 8 c}{4}

x = \frac{- ( - 4 ) - 16 - 8 c}{2 \cdot 2} : \frac{4 - 16 - 8 c}{4}

The solutions to the quadratic equation are:

x = \frac{4 + 16 - 8 c}{4}, x = \frac{4 - 16 - 8 c}{4}

4 x^{2} + 8 x = - 1

h^{2} + 10 h + 9 = 0

4 y^{2} = 4

x, f (x + 2) = 6 (x + 2)^{2} - 7

x, x^{2} = 2 y + 4 z

What are the solutions to the equation 2x^2-4x+c=0 ?
The solutions to the equation 2x^2-4x+c=0 are x=(4+sqrt(16-8c))/4 ,x=(4-sqrt(16-8c))/4
Find the zeros of 2x^2-4x+c=0
The zeros of 2x^2-4x+c=0 are x=(4+sqrt(16-8c))/4 ,x=(4-sqrt(16-8c))/4