What are the solutions to the equation 5t^2-5t-1=0 ?

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

Find the zeros of 5t^2-5t-1=0

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

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5 t^{2} - 5 t - 1 = 0

t = \frac{5 + 3}{2 5}, t = \frac{5 - 3}{2 5}

Decimal Notation

t = 1.17082 \dots, t = - 0.17082 \dots

Solution steps

5 t^{2} - 5 t - 1 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 5 ( - 1 )}{2 \cdot 5}

(- 5)^{2} - 4 \cdot 5 (- 1) = 35

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

t = \frac{5 + 3}{2 5}, t = \frac{5 - 3}{2 5}

k^{2} - 7 k + 12.25

\frac{a - y}{w + n} \cdot \frac{w ^{2} - n ^{2}}{y - a}

\frac{- 2 x + 4}{3 x - 9} \leq \frac{x + 5}{3 x - 9}

x^{2} - 3 x - 18 > 0

- 2 n \leq - 12

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