What are the solutions to the equation 2y^2+19y+24=0 ?

The solutions to the equation 2y^2+19y+24=0 are y=-3/2 ,y=-8

Find the zeros of 2y^2+19y+24=0

The zeros of 2y^2+19y+24=0 are y=-3/2 ,y=-8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 y^{2} + 19 y + 24 = 0

y = - \frac{3}{2}, y = - 8

Solution steps

2 y^{2} + 19 y + 24 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 19 \pm 1 9 ^{2} - 4 \cdot 2 \cdot 24}{2 \cdot 2}

1 9^{2} - 4 \cdot 2 \cdot 24 = 13

y_{1, 2} = \frac{- 19 \pm 13}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- 19 + 13}{2 \cdot 2}, y_{2} = \frac{- 19 - 13}{2 \cdot 2}

y = \frac{- 19 + 13}{2 \cdot 2} : - \frac{3}{2}

y = \frac{- 19 - 13}{2 \cdot 2} : - 8

The solutions to the quadratic equation are:

y = - \frac{3}{2}, y = - 8

- (x + 9)^{2} = 64

2 x - x^{2} + 3

v^{2} = 361

10 x^{2} = 9 x

4 x^{2} + x - 2 = 0

What are the solutions to the equation 2y^2+19y+24=0 ?
The solutions to the equation 2y^2+19y+24=0 are y=-3/2 ,y=-8
Find the zeros of 2y^2+19y+24=0
The zeros of 2y^2+19y+24=0 are y=-3/2 ,y=-8