What are the solutions to the equation λ^2+4λ-5(1)=0 ?

The solutions to the equation λ^2+4λ-5(1)=0 are λ=1,λ=-5

Find the zeros of λ^2+4λ-5(1)=0

The zeros of λ^2+4λ-5(1)=0 are λ=1,λ=-5

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λ^{2} + 4 λ - 5 (1) = 0

λ = 1, λ = - 5

Solution steps

λ^{2} + 4 λ - 5 (1) = 0

Expand

λ^{2} + 4 λ - 5 (1) : λ^{2} + 4 λ - 5

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}

4^{2} - 4 \cdot 1 \cdot (- 5) = 6

λ_{1, 2} = \frac{- 4 \pm 6}{2 \cdot 1}

Separate the solutions

λ_{1} = \frac{- 4 + 6}{2 \cdot 1}, λ_{2} = \frac{- 4 - 6}{2 \cdot 1}

λ = \frac{- 4 + 6}{2 \cdot 1} : 1

λ = \frac{- 4 - 6}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

λ = 1, λ = - 5

x^{2} - 16 x + 16 = 2 x - 20

2 x (x - 14) + 11 (3 + x) = - 22 x + 27 + x^{2}

- 2 x^{2} - 4 x + 30 = 0

t, y = v t + \frac{1}{2} g t^{2}

4 m^{2} + 200 m - 68696.06 = 0

What are the solutions to the equation λ^2+4λ-5(1)=0 ?
The solutions to the equation λ^2+4λ-5(1)=0 are λ=1,λ=-5
Find the zeros of λ^2+4λ-5(1)=0
The zeros of λ^2+4λ-5(1)=0 are λ=1,λ=-5