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

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

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

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

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t^{2} - 2 t = 5

t = 1 + 6, t = 1 - 6

Decimal Notation

t = 3.44948 \dots, t = - 1.44948 \dots

Solution steps

t^{2} - 2 t = 5

Move

5

to the left side

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 1 \cdot (- 5) = 26

t_{1, 2} = \frac{- ( - 2 ) \pm 2 6}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- ( - 2 ) + 2 6}{2 \cdot 1}, t_{2} = \frac{- ( - 2 ) - 2 6}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

t = 1 + 6, t = 1 - 6

x + x^{2} = 90

0 = 2 x^{2} - x + 1

5 x^{2} - 8 x + 8 = 0

y^{2} - 3 y = 2

- x^{2} + 3 = x^{2} + 1

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