What are the solutions to the equation 28x^2-4x=0 ?

The solutions to the equation 28x^2-4x=0 are x= 1/7 ,x=0

Find the zeros of 28x^2-4x=0

The zeros of 28x^2-4x=0 are x= 1/7 ,x=0

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28 x^{2} - 4 x = 0

x = \frac{1}{7}, x = 0

Decimal Notation

x = 0.14285 \dots, x = 0

Solution steps

28 x^{2} - 4 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 28 \cdot 0}{2 \cdot 28}

(- 4)^{2} - 4 \cdot 28 \cdot 0 = 4

x_{1, 2} = \frac{- ( - 4 ) \pm 4}{2 \cdot 28}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 4}{2 \cdot 28}, x_{2} = \frac{- ( - 4 ) - 4}{2 \cdot 28}

x = \frac{- ( - 4 ) + 4}{2 \cdot 28} : \frac{1}{7}

x = \frac{- ( - 4 ) - 4}{2 \cdot 28} : 0

The solutions to the quadratic equation are:

x = \frac{1}{7}, x = 0

y^{2} - 5 = 8 y

x, 3 x^{2} + 2 y^{2} = c

- 12 r^{2} - 2 = - 4 r

0.825 x^{2} + 0.15 x - 0.003621753 = 0

x, f (x + c) = 8 (x + c)^{2} + 4

What are the solutions to the equation 28x^2-4x=0 ?
The solutions to the equation 28x^2-4x=0 are x= 1/7 ,x=0
Find the zeros of 28x^2-4x=0
The zeros of 28x^2-4x=0 are x= 1/7 ,x=0