What are the solutions to the equation 2y^2+10y=-11 ?

The solutions to the equation 2y^2+10y=-11 are y=(-5+sqrt(3))/2 ,y=-(5+sqrt(3))/2

Find the zeros of 2y^2+10y=-11

The zeros of 2y^2+10y=-11 are y=(-5+sqrt(3))/2 ,y=-(5+sqrt(3))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 y^{2} + 10 y = - 11

y = \frac{- 5 + 3}{2}, y = - \frac{5 + 3}{2}

Decimal Notation

y = - 1.63397 \dots, y = - 3.36602 \dots

Solution steps

2 y^{2} + 10 y = - 11

Move

11

to the left side

2 y^{2} + 10 y + 11 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 2 \cdot 11}{2 \cdot 2}

1 0^{2} - 4 \cdot 2 \cdot 11 = 23

y_{1, 2} = \frac{- 10 \pm 2 3}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- 10 + 2 3}{2 \cdot 2}, y_{2} = \frac{- 10 - 2 3}{2 \cdot 2}

y = \frac{- 10 + 2 3}{2 \cdot 2} : \frac{- 5 + 3}{2}

y = \frac{- 10 - 2 3}{2 \cdot 2} : - \frac{5 + 3}{2}

The solutions to the quadratic equation are:

y = \frac{- 5 + 3}{2}, y = - \frac{5 + 3}{2}

16 x^{2} - 1 = 0

7 + 4 (2 a + 15) = - 13

∣ y ∣ > 6

a^{2} + ab

2 x^{4} + 4 x^{3} + 6 x^{2}

What are the solutions to the equation 2y^2+10y=-11 ?
The solutions to the equation 2y^2+10y=-11 are y=(-5+sqrt(3))/2 ,y=-(5+sqrt(3))/2
Find the zeros of 2y^2+10y=-11
The zeros of 2y^2+10y=-11 are y=(-5+sqrt(3))/2 ,y=-(5+sqrt(3))/2