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+i,t=1-i

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

The zeros of t^2-2t+2=0 are t=1+i,t=1-i

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

t = 1 + i, t = 1 - i

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}

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

t = 1 + i, t = 1 - i

(a^{\frac{2}{3}} b^{- \frac{1}{2}})^{- 6}

5 - 2 (1 - x) = 2 x - 3

\frac{( 1 + i ) ( 1 - i ) ^{- 8}}{( - 9 + 3 i 3 ) ^{- 2}}

∣ x - 1 ∣ - 4 \leq 5

0 \leq x + 2 (4 x - 1) \leq 6

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+i,t=1-i
Find the zeros of t^2-2t+2=0
The zeros of t^2-2t+2=0 are t=1+i,t=1-i