What are the solutions to the equation λ^2+λ-6(1)=0 ?

The solutions to the equation λ^2+λ-6(1)=0 are λ=2,λ=-3

Find the zeros of λ^2+λ-6(1)=0

The zeros of λ^2+λ-6(1)=0 are λ=2,λ=-3

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λ^{2} + λ - 6 (1) = 0

λ = 2, λ = - 3

Solution steps

λ^{2} + λ - 6 (1) = 0

Expand

λ^{2} + λ - 6 (1) : λ^{2} + λ - 6

λ^{2} + λ - 6 = 0

Solve with the quadratic formula

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

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

λ_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 1}

Separate the solutions

λ_{1} = \frac{- 1 + 5}{2 \cdot 1}, λ_{2} = \frac{- 1 - 5}{2 \cdot 1}

λ = \frac{- 1 + 5}{2 \cdot 1} : 2

λ = \frac{- 1 - 5}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

λ = 2, λ = - 3

(2 s^{2} + 8 s + 16) = 0

2.7 \cdot 1 0^{- 9} \cdot x^{2} - 16 \cdot 1 0^{- 5} \cdot x + 0.5 = 0

2 x^{2} = 18 x - 16

2 x^{2} - x = + 21

- 3 x^{2} + 6 x - 2 = 0.75

What are the solutions to the equation λ^2+λ-6(1)=0 ?
The solutions to the equation λ^2+λ-6(1)=0 are λ=2,λ=-3
Find the zeros of λ^2+λ-6(1)=0
The zeros of λ^2+λ-6(1)=0 are λ=2,λ=-3