What are the solutions to the equation 48=-16t^2+40t ?

The solutions to the equation 48=-16t^2+40t are t= 5/4-i(sqrt(23))/4 ,t= 5/4+i(sqrt(23))/4

Find the zeros of 48=-16t^2+40t

The zeros of 48=-16t^2+40t are t= 5/4-i(sqrt(23))/4 ,t= 5/4+i(sqrt(23))/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

48 = - 16 t^{2} + 40 t

t = \frac{5}{4} - i \frac{23}{4}, t = \frac{5}{4} + i \frac{23}{4}

Solution steps

48 = - 16 t^{2} + 40 t

Switch sides

- 16 t^{2} + 40 t = 48

Move

48

to the left side

- 16 t^{2} + 40 t - 48 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 ( - 16 ) ( - 48 )}{2 ( - 16 )}

Simplify

4 0^{2} - 4 (- 16) (- 48) : 823 i

t_{1, 2} = \frac{- 40 \pm 8 23 i}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 40 + 8 23 i}{2 ( - 16 )}, t_{2} = \frac{- 40 - 8 23 i}{2 ( - 16 )}

t = \frac{- 40 + 8 23 i}{2 ( - 16 )} : \frac{5}{4} - i \frac{23}{4}

t = \frac{- 40 - 8 23 i}{2 ( - 16 )} : \frac{5}{4} + i \frac{23}{4}

The solutions to the quadratic equation are:

t = \frac{5}{4} - i \frac{23}{4}, t = \frac{5}{4} + i \frac{23}{4}

- x^{2} + 3 x + 9 = 0

x^{2} + \frac{1}{7} x - \frac{1}{7} = 0

8 n^{2} + 47 n + 50 = - 3 n

[3 (x + 10)]^{2} + [3 x]^{2} = 20 0^{2}

\frac{3 x ^{2}}{8} = x - \frac{12}{24}

What are the solutions to the equation 48=-16t^2+40t ?
The solutions to the equation 48=-16t^2+40t are t= 5/4-i(sqrt(23))/4 ,t= 5/4+i(sqrt(23))/4
Find the zeros of 48=-16t^2+40t
The zeros of 48=-16t^2+40t are t= 5/4-i(sqrt(23))/4 ,t= 5/4+i(sqrt(23))/4