What are the solutions to the equation 12y^2+19y+5=0 ?

The solutions to the equation 12y^2+19y+5=0 are y=-1/3 ,y=-5/4

Find the zeros of 12y^2+19y+5=0

The zeros of 12y^2+19y+5=0 are y=-1/3 ,y=-5/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

12 y^{2} + 19 y + 5 = 0

y = - \frac{1}{3}, y = - \frac{5}{4}

Decimal

y = - 0.33333 \dots, y = - 1.25

Solution steps

12 y^{2} + 19 y + 5 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 19 \pm 1 9 ^{2} - 4 \cdot 12 \cdot 5}{2 \cdot 12}

1 9^{2} - 4 \cdot 12 \cdot 5 = 11

y_{1, 2} = \frac{- 19 \pm 11}{2 \cdot 12}

Separate the solutions

y_{1} = \frac{- 19 + 11}{2 \cdot 12}, y_{2} = \frac{- 19 - 11}{2 \cdot 12}

y = \frac{- 19 + 11}{2 \cdot 12} : - \frac{1}{3}

y = \frac{- 19 - 11}{2 \cdot 12} : - \frac{5}{4}

The solutions to the quadratic equation are:

y = - \frac{1}{3}, y = - \frac{5}{4}

x^{2} = - 5 x

(x^{2} + x + 3) = 0

x^{2} + 9 x + 20 = - 3 x

- x^{2} - 3 x + 2 = x + 6

y^{2} + 4 k y + 5

What are the solutions to the equation 12y^2+19y+5=0 ?
The solutions to the equation 12y^2+19y+5=0 are y=-1/3 ,y=-5/4
Find the zeros of 12y^2+19y+5=0
The zeros of 12y^2+19y+5=0 are y=-1/3 ,y=-5/4