What are the solutions to the equation 49x^2-28x+4=0 ?

The solutions to the equation 49x^2-28x+4=0 are x= 2/7

Find the zeros of 49x^2-28x+4=0

The zeros of 49x^2-28x+4=0 are x= 2/7

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49 x^{2} - 28 x + 4 = 0

x = \frac{2}{7}

Decimal Notation

x = 0.28571 \dots

Solution steps

49 x^{2} - 28 x + 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 28 ) \pm ( - 28 ) ^{2} - 4 \cdot 49 \cdot 4}{2 \cdot 49}

(- 28)^{2} - 4 \cdot 49 \cdot 4 = 0

x_{1, 2} = \frac{- ( - 28 ) \pm 0}{2 \cdot 49}

x = \frac{- ( - 28 )}{2 \cdot 49}

\frac{- ( - 28 )}{2 \cdot 49} = \frac{2}{7}

x = \frac{2}{7}

The solution to the quadratic equation is:

x = \frac{2}{7}

∣ d ∣ > 2

- 2 n - 4 = 32

- \frac{1}{4} x^{2} + x = 0

132.624 s + 1787.81 = 0

(1 + \frac{0.08}{2})^{2} - 1

What are the solutions to the equation 49x^2-28x+4=0 ?
The solutions to the equation 49x^2-28x+4=0 are x= 2/7
Find the zeros of 49x^2-28x+4=0
The zeros of 49x^2-28x+4=0 are x= 2/7