What are the solutions to the equation 49t^2-14t+1=0 ?

The solutions to the equation 49t^2-14t+1=0 are t= 1/7

Find the zeros of 49t^2-14t+1=0

The zeros of 49t^2-14t+1=0 are t= 1/7

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49 t^{2} - 14 t + 1 = 0

t = \frac{1}{7}

Decimal Notation

t = 0.14285 \dots

Solution steps

49 t^{2} - 14 t + 1 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 49 \cdot 1}{2 \cdot 49}

(- 14)^{2} - 4 \cdot 49 \cdot 1 = 0

t_{1, 2} = \frac{- ( - 14 ) \pm 0}{2 \cdot 49}

t = \frac{- ( - 14 )}{2 \cdot 49}

\frac{- ( - 14 )}{2 \cdot 49} = \frac{1}{7}

t = \frac{1}{7}

The solution to the quadratic equation is:

t = \frac{1}{7}

- 168 > - 12 a

- 5 f - 8 = 2 (3 f + 2)

\frac{9}{64} = c^{2}

1 - 3 x = 2 x - 9

0 < ∣ x - 5 ∣ < \frac{1}{2}

What are the solutions to the equation 49t^2-14t+1=0 ?
The solutions to the equation 49t^2-14t+1=0 are t= 1/7
Find the zeros of 49t^2-14t+1=0
The zeros of 49t^2-14t+1=0 are t= 1/7