What are the solutions to the equation -16t^2+48t+160=0 ?

The solutions to the equation -16t^2+48t+160=0 are t=-2,t=5

Find the zeros of -16t^2+48t+160=0

The zeros of -16t^2+48t+160=0 are t=-2,t=5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

- 16 t^{2} + 48 t + 160 = 0

t = - 2, t = 5

Solution steps

- 16 t^{2} + 48 t + 160 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 48 \pm 4 8 ^{2} - 4 ( - 16 ) \cdot 160}{2 ( - 16 )}

4 8^{2} - 4 (- 16) \cdot 160 = 112

t_{1, 2} = \frac{- 48 \pm 112}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 48 + 112}{2 ( - 16 )}, t_{2} = \frac{- 48 - 112}{2 ( - 16 )}

t = \frac{- 48 + 112}{2 ( - 16 )} : - 2

t = \frac{- 48 - 112}{2 ( - 16 )} : 5

The solutions to the quadratic equation are:

t = - 2, t = 5

0 = 4 x^{2}

2 b^{2} + 6 b - 1 = 2

x^{2} + 6 x + 11

- x^{2} - 25 = 0

x^{2} - x - 8

What are the solutions to the equation -16t^2+48t+160=0 ?
The solutions to the equation -16t^2+48t+160=0 are t=-2,t=5
Find the zeros of -16t^2+48t+160=0
The zeros of -16t^2+48t+160=0 are t=-2,t=5