What are the solutions to the equation y^2-4(-2)-4y-12=0 ?

The solutions to the equation y^2-4(-2)-4y-12=0 are y=2(1+sqrt(2)),y=2(1-sqrt(2))

Find the zeros of y^2-4(-2)-4y-12=0

The zeros of y^2-4(-2)-4y-12=0 are y=2(1+sqrt(2)),y=2(1-sqrt(2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{2} - 4 (- 2) - 4 y - 12 = 0

y = 2 (1 + 2), y = 2 (1 - 2)

Solution steps

y^{2} - 4 (- 2) - 4 y - 12 = 0

Expand

y^{2} - 4 (- 2) - 4 y - 12 : y^{2} - 4 y - 4

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

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot (- 4) = 42

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = 2 (1 + 2), y = 2 (1 - 2)

- 4 + x = - 11

- 7 < 2 x + 4 < 8

7 x = 21

2 t^{2} + 1 = 0

x^{3} + 2 x^{2} - 3 x

What are the solutions to the equation y^2-4(-2)-4y-12=0 ?
The solutions to the equation y^2-4(-2)-4y-12=0 are y=2(1+sqrt(2)),y=2(1-sqrt(2))
Find the zeros of y^2-4(-2)-4y-12=0
The zeros of y^2-4(-2)-4y-12=0 are y=2(1+sqrt(2)),y=2(1-sqrt(2))