What are the solutions to the equation y(2y+3)=5 ?

The solutions to the equation y(2y+3)=5 are y=1,y=-5/2

Find the zeros of y(2y+3)=5

The zeros of y(2y+3)=5 are y=1,y=-5/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y (2 y + 3) = 5

y = 1, y = - \frac{5}{2}

Solution steps

y (2 y + 3) = 5

Expand

y (2 y + 3) : 2 y^{2} + 3 y

2 y^{2} + 3 y = 5

Move

5

to the left side

2 y^{2} + 3 y - 5 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 ( - 5 )}{2 \cdot 2}

3^{2} - 4 \cdot 2 (- 5) = 7

y_{1, 2} = \frac{- 3 \pm 7}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- 3 + 7}{2 \cdot 2}, y_{2} = \frac{- 3 - 7}{2 \cdot 2}

y = \frac{- 3 + 7}{2 \cdot 2} : 1

y = \frac{- 3 - 7}{2 \cdot 2} : - \frac{5}{2}

The solutions to the quadratic equation are:

y = 1, y = - \frac{5}{2}

- 10 - 4 x^{2} = - 46

- 5 + 2 x^{2} = - 7 x + x^{2} - 4

5 w^{2} + 10 w = 0

- 3 x^{2} - 9 = 0

(x - 3) (x + 2) + 9 x = 3 (x^{2} - 5) - 1

What are the solutions to the equation y(2y+3)=5 ?
The solutions to the equation y(2y+3)=5 are y=1,y=-5/2
Find the zeros of y(2y+3)=5
The zeros of y(2y+3)=5 are y=1,y=-5/2