What are the solutions to the equation 2y^2+16y=-8 ?

The solutions to the equation 2y^2+16y=-8 are y=2(sqrt(3)-2),y=-2(2+sqrt(3))

Find the zeros of 2y^2+16y=-8

The zeros of 2y^2+16y=-8 are y=2(sqrt(3)-2),y=-2(2+sqrt(3))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 y^{2} + 16 y = - 8

y = 2 (3 - 2), y = - 2 (2 + 3)

Decimal Notation

y = - 0.53589 \dots, y = - 7.46410 \dots

Solution steps

2 y^{2} + 16 y = - 8

Move

8

to the left side

2 y^{2} + 16 y + 8 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 \cdot 2 \cdot 8}{2 \cdot 2}

1 6^{2} - 4 \cdot 2 \cdot 8 = 83

y_{1, 2} = \frac{- 16 \pm 8 3}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- 16 + 8 3}{2 \cdot 2}, y_{2} = \frac{- 16 - 8 3}{2 \cdot 2}

y = \frac{- 16 + 8 3}{2 \cdot 2} : 2 (3 - 2)

y = \frac{- 16 - 8 3}{2 \cdot 2} : - 2 (2 + 3)

The solutions to the quadratic equation are:

y = 2 (3 - 2), y = - 2 (2 + 3)

(x - 4)^{2} + (x - 7)^{2} = 15

y, y^{2} (2 - x) = x^{3}

4 x^{2} + 44 x + 121 = 0

16 x^{2} + 25 x + 9 = 0

x^{2} + 25 = 34

What are the solutions to the equation 2y^2+16y=-8 ?
The solutions to the equation 2y^2+16y=-8 are y=2(sqrt(3)-2),y=-2(2+sqrt(3))
Find the zeros of 2y^2+16y=-8
The zeros of 2y^2+16y=-8 are y=2(sqrt(3)-2),y=-2(2+sqrt(3))