What are the solutions to the equation 9t^2-6t+1=0 ?

The solutions to the equation 9t^2-6t+1=0 are t= 1/3

Find the zeros of 9t^2-6t+1=0

The zeros of 9t^2-6t+1=0 are t= 1/3

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9 t^{2} - 6 t + 1 = 0

t = \frac{1}{3}

Solution steps

9 t^{2} - 6 t + 1 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 9 \cdot 1}{2 \cdot 9}

(- 6)^{2} - 4 \cdot 9 \cdot 1 = 0

t_{1, 2} = \frac{- ( - 6 ) \pm 0}{2 \cdot 9}

t = \frac{- ( - 6 )}{2 \cdot 9}

\frac{- ( - 6 )}{2 \cdot 9} = \frac{1}{3}

t = \frac{1}{3}

The solution to the quadratic equation is:

t = \frac{1}{3}

18 n^{2} - 2

7 (2 x - 1) > 3 (2 x - 2)

- 5 x > 3

5 v - 8 = - 29

\frac{4}{3} π (4)^{3}

What are the solutions to the equation 9t^2-6t+1=0 ?
The solutions to the equation 9t^2-6t+1=0 are t= 1/3
Find the zeros of 9t^2-6t+1=0
The zeros of 9t^2-6t+1=0 are t= 1/3