What are the solutions to the equation 0=-16t^2+12t+40 ?

The solutions to the equation 0=-16t^2+12t+40 are t=-5/4 ,t=2

Find the zeros of 0=-16t^2+12t+40

The zeros of 0=-16t^2+12t+40 are t=-5/4 ,t=2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

0 = - 16 t^{2} + 12 t + 40

t = - \frac{5}{4}, t = 2

Decimal Notation

t = - 1.25, t = 2

Solution steps

0 = - 16 t^{2} + 12 t + 40

Switch sides

- 16 t^{2} + 12 t + 40 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 ( - 16 ) \cdot 40}{2 ( - 16 )}

1 2^{2} - 4 (- 16) \cdot 40 = 52

t_{1, 2} = \frac{- 12 \pm 52}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 12 + 52}{2 ( - 16 )}, t_{2} = \frac{- 12 - 52}{2 ( - 16 )}

t = \frac{- 12 + 52}{2 ( - 16 )} : - \frac{5}{4}

t = \frac{- 12 - 52}{2 ( - 16 )} : 2

The solutions to the quadratic equation are:

t = - \frac{5}{4}, t = 2

\frac{768 1 ^{2} \cdot 6675 \cdot 1 0 ^{3}}{6.67428 \cdot 1 0 ^{- 11}}

(y^{5})^{5}

3 ∣ x - 5 ∣ > 6

2 (a - 5) = 6

24 x + 12 y - 16 z

What are the solutions to the equation 0=-16t^2+12t+40 ?
The solutions to the equation 0=-16t^2+12t+40 are t=-5/4 ,t=2
Find the zeros of 0=-16t^2+12t+40
The zeros of 0=-16t^2+12t+40 are t=-5/4 ,t=2