What are the solutions to the equation 5=-0.04t^2+0.8t+10 ?

The solutions to the equation 5=-0.04t^2+0.8t+10 are t=-5,t=25

Find the zeros of 5=-0.04t^2+0.8t+10

The zeros of 5=-0.04t^2+0.8t+10 are t=-5,t=25

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

5 = - 0.04 t^{2} + 0.8 t + 10

t = - 5, t = 25

Solution steps

5 = - 0.04 t^{2} + 0.8 t + 10

Multiply both sides by

100

500 = - 4 t^{2} + 80 t + 1000

Switch sides

- 4 t^{2} + 80 t + 1000 = 500

Move

500

to the left side

- 4 t^{2} + 80 t + 500 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 80 \pm 8 0 ^{2} - 4 ( - 4 ) \cdot 500}{2 ( - 4 )}

8 0^{2} - 4 (- 4) \cdot 500 = 120

t_{1, 2} = \frac{- 80 \pm 120}{2 ( - 4 )}

Separate the solutions

t_{1} = \frac{- 80 + 120}{2 ( - 4 )}, t_{2} = \frac{- 80 - 120}{2 ( - 4 )}

t = \frac{- 80 + 120}{2 ( - 4 )} : - 5

t = \frac{- 80 - 120}{2 ( - 4 )} : 25

The solutions to the quadratic equation are:

t = - 5, t = 25

x, z^{2} = 9 x^{2} + y^{2}

y, 3 y^{2} + 2 a y = - b

ab x^{2} + ab = a^{2} x + b^{2} x

x^{2} + 6 x = - 8

12 x^{2} - 19 x + 5 = 0

What are the solutions to the equation 5=-0.04t^2+0.8t+10 ?
The solutions to the equation 5=-0.04t^2+0.8t+10 are t=-5,t=25
Find the zeros of 5=-0.04t^2+0.8t+10
The zeros of 5=-0.04t^2+0.8t+10 are t=-5,t=25