What are the solutions to the equation -3λ+λ^2+2=0 ?

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

Find the zeros of -3λ+λ^2+2=0

The zeros of -3λ+λ^2+2=0 are λ=2,λ=1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

- 3 λ + λ^{2} + 2 = 0

λ = 2, λ = 1

Solution steps

- 3 λ + λ^{2} + 2 = 0

Write in the standard form

a x^{2} + b x + c = 0

λ^{2} - 3 λ + 2 = 0

Solve with the quadratic formula

λ_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot 2}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot 2 = 1

λ_{1, 2} = \frac{- ( - 3 ) \pm 1}{2 \cdot 1}

Separate the solutions

λ_{1} = \frac{- ( - 3 ) + 1}{2 \cdot 1}, λ_{2} = \frac{- ( - 3 ) - 1}{2 \cdot 1}

λ = \frac{- ( - 3 ) + 1}{2 \cdot 1} : 2

λ = \frac{- ( - 3 ) - 1}{2 \cdot 1} : 1

The solutions to the quadratic equation are:

λ = 2, λ = 1

3 n^{2} - 75 = 0

u^{2} + 18 u

(1 - 3) x^{2} - (3 + 1) x - 2 = 0

3 x^{2} + 12 = 3

5 x^{2} - x - 108 = 0

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