What are the solutions to the equation λ^2-4λ+15=0 ?

The solutions to the equation λ^2-4λ+15=0 are λ=2+sqrt(11)i,λ=2-sqrt(11)i

Find the zeros of λ^2-4λ+15=0

The zeros of λ^2-4λ+15=0 are λ=2+sqrt(11)i,λ=2-sqrt(11)i

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λ^{2} - 4 λ + 15 = 0

λ = 2 + 11 i, λ = 2 - 11 i

Solution steps

λ^{2} - 4 λ + 15 = 0

Solve with the quadratic formula

λ_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 15}{2 \cdot 1}

Simplify

(- 4)^{2} - 4 \cdot 1 \cdot 15 : 211 i

λ_{1, 2} = \frac{- ( - 4 ) \pm 2 11 i}{2 \cdot 1}

Separate the solutions

λ_{1} = \frac{- ( - 4 ) + 2 11 i}{2 \cdot 1}, λ_{2} = \frac{- ( - 4 ) - 2 11 i}{2 \cdot 1}

λ = \frac{- ( - 4 ) + 2 11 i}{2 \cdot 1} : 2 + 11 i

λ = \frac{- ( - 4 ) - 2 11 i}{2 \cdot 1} : 2 - 11 i

The solutions to the quadratic equation are:

λ = 2 + 11 i, λ = 2 - 11 i

x^{2} > 9

4 x - 2 x^{2} = \frac{4}{3}

\frac{( - 2 ) ^{7}}{( - 2 ) ^{5}}

∣ 2 x - 3 ∣ + 9 \geq 20

\frac{π}{2} - \frac{4 π}{9}

What are the solutions to the equation λ^2-4λ+15=0 ?
The solutions to the equation λ^2-4λ+15=0 are λ=2+sqrt(11)i,λ=2-sqrt(11)i
Find the zeros of λ^2-4λ+15=0
The zeros of λ^2-4λ+15=0 are λ=2+sqrt(11)i,λ=2-sqrt(11)i