What are the solutions to the equation 12t^2-10t+5=0 ?

The solutions to the equation 12t^2-10t+5=0 are t= 5/12+i(sqrt(35))/(12),t= 5/12-i(sqrt(35))/(12)

Find the zeros of 12t^2-10t+5=0

The zeros of 12t^2-10t+5=0 are t= 5/12+i(sqrt(35))/(12),t= 5/12-i(sqrt(35))/(12)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

12 t^{2} - 10 t + 5 = 0

t = \frac{5}{12} + i \frac{35}{12}, t = \frac{5}{12} - i \frac{35}{12}

Solution steps

12 t^{2} - 10 t + 5 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 12 \cdot 5}{2 \cdot 12}

Simplify

(- 10)^{2} - 4 \cdot 12 \cdot 5 : 235 i

t_{1, 2} = \frac{- ( - 10 ) \pm 2 35 i}{2 \cdot 12}

Separate the solutions

t_{1} = \frac{- ( - 10 ) + 2 35 i}{2 \cdot 12}, t_{2} = \frac{- ( - 10 ) - 2 35 i}{2 \cdot 12}

t = \frac{- ( - 10 ) + 2 35 i}{2 \cdot 12} : \frac{5}{12} + i \frac{35}{12}

t = \frac{- ( - 10 ) - 2 35 i}{2 \cdot 12} : \frac{5}{12} - i \frac{35}{12}

The solutions to the quadratic equation are:

t = \frac{5}{12} + i \frac{35}{12}, t = \frac{5}{12} - i \frac{35}{12}

2 x^{2} + x + 0 = 0

1 - 8 x + 16 x^{2} = 2 - 8 x

x, x^{2} - 2 x - 35 = 0

6 x^{2} - 7 = 8 x

- 4 (x - 3)^{2} = - 20

What are the solutions to the equation 12t^2-10t+5=0 ?
The solutions to the equation 12t^2-10t+5=0 are t= 5/12+i(sqrt(35))/(12),t= 5/12-i(sqrt(35))/(12)
Find the zeros of 12t^2-10t+5=0
The zeros of 12t^2-10t+5=0 are t= 5/12+i(sqrt(35))/(12),t= 5/12-i(sqrt(35))/(12)