What are the solutions to the equation 5t^2-4t=11 ?

The solutions to the equation 5t^2-4t=11 are t=(2+sqrt(59))/5 ,t=(2-sqrt(59))/5

Find the zeros of 5t^2-4t=11

The zeros of 5t^2-4t=11 are t=(2+sqrt(59))/5 ,t=(2-sqrt(59))/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

5 t^{2} - 4 t = 11

t = \frac{2 + 59}{5}, t = \frac{2 - 59}{5}

Decimal Notation

t = 1.93622 \dots, t = - 1.13622 \dots

Solution steps

5 t^{2} - 4 t = 11

Move

11

to the left side

5 t^{2} - 4 t - 11 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 5 ( - 11 )}{2 \cdot 5}

(- 4)^{2} - 4 \cdot 5 (- 11) = 259

t_{1, 2} = \frac{- ( - 4 ) \pm 2 59}{2 \cdot 5}

Separate the solutions

t_{1} = \frac{- ( - 4 ) + 2 59}{2 \cdot 5}, t_{2} = \frac{- ( - 4 ) - 2 59}{2 \cdot 5}

t = \frac{- ( - 4 ) + 2 59}{2 \cdot 5} : \frac{2 + 59}{5}

t = \frac{- ( - 4 ) - 2 59}{2 \cdot 5} : \frac{2 - 59}{5}

The solutions to the quadratic equation are:

t = \frac{2 + 59}{5}, t = \frac{2 - 59}{5}

x^{2} + 3 x = 270

7 x^{2} - 28 x + 17 = - 4

2 x^{2} + 14 = 16 x

x^{2} + x - 7 = 5

s^{2} + 4 s + 12 = 0

What are the solutions to the equation 5t^2-4t=11 ?
The solutions to the equation 5t^2-4t=11 are t=(2+sqrt(59))/5 ,t=(2-sqrt(59))/5
Find the zeros of 5t^2-4t=11
The zeros of 5t^2-4t=11 are t=(2+sqrt(59))/5 ,t=(2-sqrt(59))/5