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

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

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

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

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

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

Decimal Notation

t = 3.61803 \dots, t = 1.38196 \dots

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

x, k x^{2} - 17 x + 33 = 0

1 2^{2} = m \cdot m + 7

8 x^{2} + \frac{8}{4} x + \frac{8}{5} = 0

\frac{m - 1}{2} = (m - 1) (m + 3)

32 = 2 x^{2}

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