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

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

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

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

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

t = 1 + 3, t = 1 - 3

Solution steps

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 1 \cdot (- 2) = 23

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

t = 1 + 3, t = 1 - 3

2 k^{2} + 9 k = - 7

\frac{1}{4} - \frac{1}{3} x = \frac{1}{3}

\frac{∣ x + 2 ∣}{3} = - 3

(2 x - 1)^{2} = 9

9 x^{4} - 16 y^{4}

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