What are the solutions to the equation 9t^2-3t-1=0 ?

The solutions to the equation 9t^2-3t-1=0 are t=(1+sqrt(5))/6 ,t=(1-sqrt(5))/6

Find the zeros of 9t^2-3t-1=0

The zeros of 9t^2-3t-1=0 are t=(1+sqrt(5))/6 ,t=(1-sqrt(5))/6

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9 t^{2} - 3 t - 1 = 0

t = \frac{1 + 5}{6}, t = \frac{1 - 5}{6}

Decimal Notation

t = 0.53934 \dots, t = - 0.20601 \dots

Solution steps

9 t^{2} - 3 t - 1 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 9 ( - 1 )}{2 \cdot 9}

(- 3)^{2} - 4 \cdot 9 (- 1) = 35

t_{1, 2} = \frac{- ( - 3 ) \pm 3 5}{2 \cdot 9}

Separate the solutions

t_{1} = \frac{- ( - 3 ) + 3 5}{2 \cdot 9}, t_{2} = \frac{- ( - 3 ) - 3 5}{2 \cdot 9}

t = \frac{- ( - 3 ) + 3 5}{2 \cdot 9} : \frac{1 + 5}{6}

t = \frac{- ( - 3 ) - 3 5}{2 \cdot 9} : \frac{1 - 5}{6}

The solutions to the quadratic equation are:

t = \frac{1 + 5}{6}, t = \frac{1 - 5}{6}

x^{4} - x^{2} - 90

2 t^{2} + 11 t + 9

\frac{( x ^{2} y ^{3} ) ^{4} ( x y ^{4} ) ^{- 3}}{x ^{2} y}

\frac{m + 8}{- 2} = - 5

- 15 - 15

What are the solutions to the equation 9t^2-3t-1=0 ?
The solutions to the equation 9t^2-3t-1=0 are t=(1+sqrt(5))/6 ,t=(1-sqrt(5))/6
Find the zeros of 9t^2-3t-1=0
The zeros of 9t^2-3t-1=0 are t=(1+sqrt(5))/6 ,t=(1-sqrt(5))/6