What are the solutions to the equation t^2-2t+5=0 ?

The solutions to the equation t^2-2t+5=0 are t=1+2i,t=1-2i

Find the zeros of t^2-2t+5=0

The zeros of t^2-2t+5=0 are t=1+2i,t=1-2i

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

t = 1 + 2 i, t = 1 - 2 i

Solution steps

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}

Simplify

(- 2)^{2} - 4 \cdot 1 \cdot 5 : 4 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

t = 1 + 2 i, t = 1 - 2 i

- \frac{3}{2} x + 5 = - 4

4 y^{2} - 81

4 n^{3} + 3 n^{2} + 4 n + 3

\frac{1}{2} (- 6)^{2}

8 π = 2 π r (r + 3)

What are the solutions to the equation t^2-2t+5=0 ?
The solutions to the equation t^2-2t+5=0 are t=1+2i,t=1-2i
Find the zeros of t^2-2t+5=0
The zeros of t^2-2t+5=0 are t=1+2i,t=1-2i