What are the solutions to the equation y^2-4y-1=0 ?

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

Find the zeros of y^2-4y-1=0

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{2} - 4 y - 1 = 0

y = 2 + 5, y = 2 - 5

Decimal Notation

y = 4.23606 \dots, y = - 0.23606 \dots

Solution steps

y^{2} - 4 y - 1 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot (- 1) = 25

y_{1, 2} = \frac{- ( - 4 ) \pm 2 5}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 4 ) + 2 5}{2 \cdot 1}, y_{2} = \frac{- ( - 4 ) - 2 5}{2 \cdot 1}

y = \frac{- ( - 4 ) + 2 5}{2 \cdot 1} : 2 + 5

y = \frac{- ( - 4 ) - 2 5}{2 \cdot 1} : 2 - 5

The solutions to the quadratic equation are:

y = 2 + 5, y = 2 - 5

\frac{15 x ^{2} y ^{2} + 5 x ^{2} y - 10 x y ^{2}}{- x y}

- 16 t^{2} + 96 = 0

- 2 = - \frac{1}{4} x - \frac{1}{2}

\frac{a}{12 b} + \frac{b}{8 a}

2 (m - 3) + 7 < 21

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