What are the solutions to the equation 2y^2+2y-12=0 ?

The solutions to the equation 2y^2+2y-12=0 are y=2,y=-3

Find the zeros of 2y^2+2y-12=0

The zeros of 2y^2+2y-12=0 are y=2,y=-3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 y^{2} + 2 y - 12 = 0

y = 2, y = - 3

Solution steps

2 y^{2} + 2 y - 12 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 2 ( - 12 )}{2 \cdot 2}

2^{2} - 4 \cdot 2 (- 12) = 10

y_{1, 2} = \frac{- 2 \pm 10}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- 2 + 10}{2 \cdot 2}, y_{2} = \frac{- 2 - 10}{2 \cdot 2}

y = \frac{- 2 + 10}{2 \cdot 2} : 2

y = \frac{- 2 - 10}{2 \cdot 2} : - 3

The solutions to the quadratic equation are:

y = 2, y = - 3

3 x^{2} - x - 4 = 0

y, x = y^{2} - 4 y + 5

x, y = 2 \cdot x - x^{2}

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

- x^{2} + 4 x + 6

What are the solutions to the equation 2y^2+2y-12=0 ?
The solutions to the equation 2y^2+2y-12=0 are y=2,y=-3
Find the zeros of 2y^2+2y-12=0
The zeros of 2y^2+2y-12=0 are y=2,y=-3