What are the solutions to the equation 16y^2+62y-45=0 ?

The solutions to the equation 16y^2+62y-45=0 are y= 5/8 ,y=-9/2

Find the zeros of 16y^2+62y-45=0

The zeros of 16y^2+62y-45=0 are y= 5/8 ,y=-9/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

16 y^{2} + 62 y - 45 = 0

y = \frac{5}{8}, y = - \frac{9}{2}

Decimal Notation

y = 0.625, y = - 4.5

Solution steps

16 y^{2} + 62 y - 45 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 62 \pm 6 2 ^{2} - 4 \cdot 16 ( - 45 )}{2 \cdot 16}

6 2^{2} - 4 \cdot 16 (- 45) = 82

y_{1, 2} = \frac{- 62 \pm 82}{2 \cdot 16}

Separate the solutions

y_{1} = \frac{- 62 + 82}{2 \cdot 16}, y_{2} = \frac{- 62 - 82}{2 \cdot 16}

y = \frac{- 62 + 82}{2 \cdot 16} : \frac{5}{8}

y = \frac{- 62 - 82}{2 \cdot 16} : - \frac{9}{2}

The solutions to the quadratic equation are:

y = \frac{5}{8}, y = - \frac{9}{2}

y, y^{2} - t^{2} + 2 t y = C

24 x^{2} - 10 x - 3 = 0

14 x^{2} + 6 x - 20 = 0

x, x^{2} + 5 y^{2} - 2 x + 10 y = z^{2}

- 0.002 x^{2} + 0.4 x + 1.8

What are the solutions to the equation 16y^2+62y-45=0 ?
The solutions to the equation 16y^2+62y-45=0 are y= 5/8 ,y=-9/2
Find the zeros of 16y^2+62y-45=0
The zeros of 16y^2+62y-45=0 are y= 5/8 ,y=-9/2