What are the solutions to the equation 100t^2-192t+256=0 ?

The solutions to the equation 100t^2-192t+256=0 are t= 24/25+i 32/25 ,t= 24/25-i 32/25

Find the zeros of 100t^2-192t+256=0

The zeros of 100t^2-192t+256=0 are t= 24/25+i 32/25 ,t= 24/25-i 32/25

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

100 t^{2} - 192 t + 256 = 0

t = \frac{24}{25} + i \frac{32}{25}, t = \frac{24}{25} - i \frac{32}{25}

Solution steps

100 t^{2} - 192 t + 256 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 192 ) \pm ( - 192 ) ^{2} - 4 \cdot 100 \cdot 256}{2 \cdot 100}

Simplify

(- 192)^{2} - 4 \cdot 100 \cdot 256 : 256 i

t_{1, 2} = \frac{- ( - 192 ) \pm 256 i}{2 \cdot 100}

Separate the solutions

t_{1} = \frac{- ( - 192 ) + 256 i}{2 \cdot 100}, t_{2} = \frac{- ( - 192 ) - 256 i}{2 \cdot 100}

t = \frac{- ( - 192 ) + 256 i}{2 \cdot 100} : \frac{24}{25} + i \frac{32}{25}

t = \frac{- ( - 192 ) - 256 i}{2 \cdot 100} : \frac{24}{25} - i \frac{32}{25}

The solutions to the quadratic equation are:

t = \frac{24}{25} + i \frac{32}{25}, t = \frac{24}{25} - i \frac{32}{25}

\frac{a}{b} = k^{2}

x, j = (4 + 3 x) (8 x + s)

x, x - y = 4 y^{3} - 3 x^{2}

(4 a)^{2} + (3 a)^{2} = 12 5^{2}

- 72 = - 18 x^{2}

What are the solutions to the equation 100t^2-192t+256=0 ?
The solutions to the equation 100t^2-192t+256=0 are t= 24/25+i 32/25 ,t= 24/25-i 32/25
Find the zeros of 100t^2-192t+256=0
The zeros of 100t^2-192t+256=0 are t= 24/25+i 32/25 ,t= 24/25-i 32/25