What are the solutions to the equation t^2-4t+2=0 ?

The solutions to the equation t^2-4t+2=0 are t=2+sqrt(2),t=2-sqrt(2)

Find the zeros of t^2-4t+2=0

The zeros of t^2-4t+2=0 are t=2+sqrt(2),t=2-sqrt(2)

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t^{2} - 4 t + 2 = 0

t = 2 + 2, t = 2 - 2

Decimal Notation

t = 3.41421 \dots, t = 0.58578 \dots

Solution steps

t^{2} - 4 t + 2 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot 2 = 22

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

Separate the solutions

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

t = \frac{- ( - 4 ) + 2 2}{2 \cdot 1} : 2 + 2

t = \frac{- ( - 4 ) - 2 2}{2 \cdot 1} : 2 - 2

The solutions to the quadratic equation are:

t = 2 + 2, t = 2 - 2

e^{1.2}

f (n) = \frac{2 ^{n}}{2 ^{n + 1}}

- 4 < 2 t - 6 < 8

x^{2} - x - 46 = 3 x - 1

x^{2} - 3 x - 2 \leq x + 3

What are the solutions to the equation t^2-4t+2=0 ?
The solutions to the equation t^2-4t+2=0 are t=2+sqrt(2),t=2-sqrt(2)
Find the zeros of t^2-4t+2=0
The zeros of t^2-4t+2=0 are t=2+sqrt(2),t=2-sqrt(2)