What are the solutions to the equation (λ^2-14)-20=λ ?

The solutions to the equation (λ^2-14)-20=λ are λ=(1+sqrt(137))/2 ,λ=(1-sqrt(137))/2

Find the zeros of (λ^2-14)-20=λ

The zeros of (λ^2-14)-20=λ are λ=(1+sqrt(137))/2 ,λ=(1-sqrt(137))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(λ^{2} - 14) - 20 = λ

λ = \frac{1 + 137}{2}, λ = \frac{1 - 137}{2}

Decimal Notation

λ = 6.35234 \dots, λ = - 5.35234 \dots

Solution steps

(λ^{2} - 14) - 20 = λ

Refine

λ^{2} - 34 = λ

Move

λ

to the left side

λ^{2} - 34 - λ = 0

Write in the standard form

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

λ^{2} - λ - 34 = 0

Solve with the quadratic formula

λ_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 34 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 34) = 137

λ_{1, 2} = \frac{- ( - 1 ) \pm 137}{2 \cdot 1}

Separate the solutions

λ_{1} = \frac{- ( - 1 ) + 137}{2 \cdot 1}, λ_{2} = \frac{- ( - 1 ) - 137}{2 \cdot 1}

λ = \frac{- ( - 1 ) + 137}{2 \cdot 1} : \frac{1 + 137}{2}

λ = \frac{- ( - 1 ) - 137}{2 \cdot 1} : \frac{1 - 137}{2}

The solutions to the quadratic equation are:

λ = \frac{1 + 137}{2}, λ = \frac{1 - 137}{2}

x, x^{2} - 2 x = m^{2} + 2 m

- 3 x^{2} + (7 k^{2} - 3) x + 7 = 0

w, z = - 2 w^{2}

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

x^{2} + 3 x = 20 x

What are the solutions to the equation (λ^2-14)-20=λ ?
The solutions to the equation (λ^2-14)-20=λ are λ=(1+sqrt(137))/2 ,λ=(1-sqrt(137))/2
Find the zeros of (λ^2-14)-20=λ
The zeros of (λ^2-14)-20=λ are λ=(1+sqrt(137))/2 ,λ=(1-sqrt(137))/2