What are the solutions to the equation 120=-5t^2+40t+40 ?

The solutions to the equation 120=-5t^2+40t+40 are t=4

Find the zeros of 120=-5t^2+40t+40

The zeros of 120=-5t^2+40t+40 are t=4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

120 = - 5 t^{2} + 40 t + 40

t = 4

Solution steps

120 = - 5 t^{2} + 40 t + 40

Switch sides

- 5 t^{2} + 40 t + 40 = 120

Move

120

to the left side

- 5 t^{2} + 40 t - 80 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 ( - 5 ) ( - 80 )}{2 ( - 5 )}

4 0^{2} - 4 (- 5) (- 80) = 0

t_{1, 2} = \frac{- 40 \pm 0}{2 ( - 5 )}

t = \frac{- 40}{2 ( - 5 )}

\frac{- 40}{2 ( - 5 )} = 4

t = 4

The solution to the quadratic equation is:

t = 4

- w^{2} + 5 w - 4 = 0

x = 16 - 0.0003 (x - 3)^{2} 2200

x, f (x + 3) = x^{2} + 3

3 (x^{2} - 4 x + 4) = 0

x^{2} - 2 x + 1 = 2 x + 5

What are the solutions to the equation 120=-5t^2+40t+40 ?
The solutions to the equation 120=-5t^2+40t+40 are t=4
Find the zeros of 120=-5t^2+40t+40
The zeros of 120=-5t^2+40t+40 are t=4