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

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

Find the zeros of λ^2-4λ+5=0

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

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λ^{2} - 4 λ + 5 = 0

λ = 2 + i, λ = 2 - i

Solution steps

λ^{2} - 4 λ + 5 = 0

Solve with the quadratic formula

λ_{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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

λ = 2 + i, λ = 2 - i

3 x^{2} = 2 x + 3

4 x + 34 = 106

2 (\frac{5 - y}{3}) - 3 y = 7

r - 13 = 3

x^{2} - 8 x y - 20 y^{2}

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