What are the solutions to the equation y^2+y=3 ?

The solutions to the equation y^2+y=3 are y=(-1+sqrt(13))/2 ,y=(-1-sqrt(13))/2

Find the zeros of y^2+y=3

The zeros of y^2+y=3 are y=(-1+sqrt(13))/2 ,y=(-1-sqrt(13))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{2} + y = 3

y = \frac{- 1 + 13}{2}, y = \frac{- 1 - 13}{2}

Decimal Notation

y = 1.30277 \dots, y = - 2.30277 \dots

Solution steps

y^{2} + y = 3

Move

3

to the left side

y^{2} + y - 3 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 3 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 3) = 13

y_{1, 2} = \frac{- 1 \pm 13}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 1 + 13}{2 \cdot 1}, y_{2} = \frac{- 1 - 13}{2 \cdot 1}

y = \frac{- 1 + 13}{2 \cdot 1} : \frac{- 1 + 13}{2}

y = \frac{- 1 - 13}{2 \cdot 1} : \frac{- 1 - 13}{2}

The solutions to the quadratic equation are:

y = \frac{- 1 + 13}{2}, y = \frac{- 1 - 13}{2}

x^{2} + x - 7 - 5 = 0

(6)^{2} + (8)^{2} = c^{2}

2 (3 x - 5)^{2} - 2 = 70

3 x^{2} + 2 x = 0

0 = \frac{18 - x ^{2}}{4}

What are the solutions to the equation y^2+y=3 ?
The solutions to the equation y^2+y=3 are y=(-1+sqrt(13))/2 ,y=(-1-sqrt(13))/2
Find the zeros of y^2+y=3
The zeros of y^2+y=3 are y=(-1+sqrt(13))/2 ,y=(-1-sqrt(13))/2