What are the solutions to the equation -2(y^2+2y)+3(y^2-1)=5 ?

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

Find the zeros of -2(y^2+2y)+3(y^2-1)=5

The zeros of -2(y^2+2y)+3(y^2-1)=5 are y=2(1+sqrt(3)),y=2(1-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} + 2 y) + 3 (y^{2} - 1) = 5

y = 2 (1 + 3), y = 2 (1 - 3)

Decimal Notation

y = 5.46410 \dots, y = - 1.46410 \dots

Solution steps

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

Expand

- 2 (y^{2} + 2 y) + 3 (y^{2} - 1) : y^{2} - 4 y - 3

y^{2} - 4 y - 3 = 5

Move

5

to the left side

y^{2} - 4 y - 8 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 8) = 43

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

Separate the solutions

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

y = \frac{- ( - 4 ) + 4 3}{2 \cdot 1} : 2 (1 + 3)

y = \frac{- ( - 4 ) - 4 3}{2 \cdot 1} : 2 (1 - 3)

The solutions to the quadratic equation are:

y = 2 (1 + 3), y = 2 (1 - 3)

1 = 2 x^{2}

6 b^{2} + 18 b - 24 = 0

- x^{2} + 6 x = 3

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

x^{2} - 30 x^{2} + 125 = 0

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