What are the solutions to the equation solvefor t,t^2-2yt-1=0 ?

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

Find the zeros of solvefor t,t^2-2yt-1=0

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

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solve for

t, t^{2} - 2 y t - 1 = 0

t = y + y^{2} + 1, t = y - y^{2} + 1

Solution steps

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

Solve with the quadratic formula

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

Simplify

(- 2 y)^{2} - 4 \cdot 1 \cdot (- 1) : 2 y^{2} + 1

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

t = y + y^{2} + 1, t = y - y^{2} + 1

x^{2} - 6 x - 28 = 0

7 (7 m - 8) (m - 3) = 0

- 16 x^{2} + 128 x = 0

8.667 t^{2} - 349.67 t + 3523 = 35

2800 - 16 x - 24 x^{2} = 0

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