What are the solutions to the equation α^2-4α+13=0 ?

The solutions to the equation α^2-4α+13=0 are α=2+3i,α=2-3i

Find the zeros of α^2-4α+13=0

The zeros of α^2-4α+13=0 are α=2+3i,α=2-3i

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α^{2} - 4 α + 13 = 0

α = 2 + 3 i, α = 2 - 3 i

Solution steps

α^{2} - 4 α + 13 = 0

Solve with the quadratic formula

α_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 13}{2 \cdot 1}

Simplify

(- 4)^{2} - 4 \cdot 1 \cdot 13 : 6 i

α_{1, 2} = \frac{- ( - 4 ) \pm 6 i}{2 \cdot 1}

Separate the solutions

α_{1} = \frac{- ( - 4 ) + 6 i}{2 \cdot 1}, α_{2} = \frac{- ( - 4 ) - 6 i}{2 \cdot 1}

α = \frac{- ( - 4 ) + 6 i}{2 \cdot 1} : 2 + 3 i

α = \frac{- ( - 4 ) - 6 i}{2 \cdot 1} : 2 - 3 i

The solutions to the quadratic equation are:

α = 2 + 3 i, α = 2 - 3 i

(x - 2)^{2} = - 6

(5)^{\frac{7}{3}}

\frac{6 !}{2 ! 4 !}

\frac{1}{c ^{- 1}}

- 4 n + 11.5 = - 12.5

What are the solutions to the equation α^2-4α+13=0 ?
The solutions to the equation α^2-4α+13=0 are α=2+3i,α=2-3i
Find the zeros of α^2-4α+13=0
The zeros of α^2-4α+13=0 are α=2+3i,α=2-3i