What are the solutions to the equation y^2+y/(1.25)-5=0 ?

The solutions to the equation y^2+y/(1.25)-5=0 are y=(-1+sqrt(32.25))/(2.5),y=(-1-sqrt(32.25))/(2.5)

Find the zeros of y^2+y/(1.25)-5=0

The zeros of y^2+y/(1.25)-5=0 are y=(-1+sqrt(32.25))/(2.5),y=(-1-sqrt(32.25))/(2.5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{2} + \frac{y}{1.25} - 5 = 0

y = \frac{- 1 + 32.25}{2.5}, y = \frac{- 1 - 32.25}{2.5}

Decimal Notation

y = 1.87156 \dots, y = - 2.67156 \dots

Solution steps

y^{2} + \frac{y}{1.25} - 5 = 0

Multiply both sides by

1.25

1.25 y^{2} + y - 6.25 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1.25 ( - 6.25 )}{2 \cdot 1.25}

1^{2} - 4 \cdot 1.25 (- 6.25) = 32.25

y_{1, 2} = \frac{- 1 \pm 32.25}{2 \cdot 1.25}

Separate the solutions

y_{1} = \frac{- 1 + 32.25}{2 \cdot 1.25}, y_{2} = \frac{- 1 - 32.25}{2 \cdot 1.25}

y = \frac{- 1 + 32.25}{2 \cdot 1.25} : \frac{- 1 + 32.25}{2.5}

y = \frac{- 1 - 32.25}{2 \cdot 1.25} : \frac{- 1 - 32.25}{2.5}

The solutions to the quadratic equation are:

y = \frac{- 1 + 32.25}{2.5}, y = \frac{- 1 - 32.25}{2.5}

- 1 = x^{2} + 4

x^{2} = - 16

x^{2} + 8 x + 25 = 0

8 x^{2} + 9 = 313

16 x^{2} - 25 = 0

What are the solutions to the equation y^2+y/(1.25)-5=0 ?
The solutions to the equation y^2+y/(1.25)-5=0 are y=(-1+sqrt(32.25))/(2.5),y=(-1-sqrt(32.25))/(2.5)
Find the zeros of y^2+y/(1.25)-5=0
The zeros of y^2+y/(1.25)-5=0 are y=(-1+sqrt(32.25))/(2.5),y=(-1-sqrt(32.25))/(2.5)