What are the solutions to the equation t^2-4t+5=0 ?

The solutions to the equation t^2-4t+5=0 are t=2+i,t=2-i

Find the zeros of t^2-4t+5=0

The zeros of t^2-4t+5=0 are t=2+i,t=2-i

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t^{2} - 4 t + 5 = 0

t = 2 + i, t = 2 - i

Solution steps

t^{2} - 4 t + 5 = 0

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

t = 2 + i, t = 2 - i

\frac{x + 1}{1 - x} \geq 0

y (y + 1) - 4 = 0

2 (2 q + 1.5) = 18 - q

3 x - 3 > 34

What are the solutions to the equation t^2-4t+5=0 ?
The solutions to the equation t^2-4t+5=0 are t=2+i,t=2-i
Find the zeros of t^2-4t+5=0
The zeros of t^2-4t+5=0 are t=2+i,t=2-i