What are the solutions to the equation -3t^2+4t+40=-t+15 ?

The solutions to the equation -3t^2+4t+40=-t+15 are t=-(-5+5sqrt(13))/6 ,t=(5+5sqrt(13))/6

Find the zeros of -3t^2+4t+40=-t+15

The zeros of -3t^2+4t+40=-t+15 are t=-(-5+5sqrt(13))/6 ,t=(5+5sqrt(13))/6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

- 3 t^{2} + 4 t + 40 = - t + 15

t = - \frac{- 5 + 5 13}{6}, t = \frac{5 + 5 13}{6}

Decimal Notation

t = - 2.17129 \dots, t = 3.83795 \dots

Solution steps

- 3 t^{2} + 4 t + 40 = - t + 15

Move

15

to the left side

- 3 t^{2} + 4 t + 25 = - t

Move

t

to the left side

- 3 t^{2} + 5 t + 25 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 ( - 3 ) \cdot 25}{2 ( - 3 )}

5^{2} - 4 (- 3) \cdot 25 = 513

t_{1, 2} = \frac{- 5 \pm 5 13}{2 ( - 3 )}

Separate the solutions

t_{1} = \frac{- 5 + 5 13}{2 ( - 3 )}, t_{2} = \frac{- 5 - 5 13}{2 ( - 3 )}

t = \frac{- 5 + 5 13}{2 ( - 3 )} : - \frac{- 5 + 5 13}{6}

t = \frac{- 5 - 5 13}{2 ( - 3 )} : \frac{5 + 5 13}{6}

The solutions to the quadratic equation are:

t = - \frac{- 5 + 5 13}{6}, t = \frac{5 + 5 13}{6}

a^{2} + (\frac{5 3}{2} + a)^{2} = 5^{2}

5 n^{2} + 9.8 n - 1200 = 0

x^{2} - 3 x - 26 = 2

x, y^{3} + 3 x^{2} y = c_{1}

8 x^{2} + 7 x - 1666 = 0

What are the solutions to the equation -3t^2+4t+40=-t+15 ?
The solutions to the equation -3t^2+4t+40=-t+15 are t=-(-5+5sqrt(13))/6 ,t=(5+5sqrt(13))/6
Find the zeros of -3t^2+4t+40=-t+15
The zeros of -3t^2+4t+40=-t+15 are t=-(-5+5sqrt(13))/6 ,t=(5+5sqrt(13))/6