What are the solutions to the equation t(13-t)=40 ?

The solutions to the equation t(13-t)=40 are t=5,t=8

Find the zeros of t(13-t)=40

The zeros of t(13-t)=40 are t=5,t=8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

t (13 - t) = 40

t = 5, t = 8

Solution steps

t (13 - t) = 40

Expand

t (13 - t) : 13 t - t^{2}

13 t - t^{2} = 40

Move

40

to the left side

13 t - t^{2} - 40 = 0

Write in the standard form

a x^{2} + b x + c = 0

- t^{2} + 13 t - 40 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 ( - 1 ) ( - 40 )}{2 ( - 1 )}

1 3^{2} - 4 (- 1) (- 40) = 3

t_{1, 2} = \frac{- 13 \pm 3}{2 ( - 1 )}

Separate the solutions

t_{1} = \frac{- 13 + 3}{2 ( - 1 )}, t_{2} = \frac{- 13 - 3}{2 ( - 1 )}

t = \frac{- 13 + 3}{2 ( - 1 )} : 5

t = \frac{- 13 - 3}{2 ( - 1 )} : 8

The solutions to the quadratic equation are:

t = 5, t = 8

x, x^{2} + y^{2} - 6 x + 4 y + 13 = 0

x^{2} + 6 x - 16 = 0

x^{2} + (m + 4) x + \frac{3 m}{2} + \frac{9}{2} = 0

x, x^{2} + 11 k x + 40 = 0

(2 x - 3) (x + 6) = 0

What are the solutions to the equation t(13-t)=40 ?
The solutions to the equation t(13-t)=40 are t=5,t=8
Find the zeros of t(13-t)=40
The zeros of t(13-t)=40 are t=5,t=8