What are the solutions to the equation 8x^2-4x+5=0 ?

The solutions to the equation 8x^2-4x+5=0 are x= 1/4+i 3/4 ,x= 1/4-i 3/4

Find the zeros of 8x^2-4x+5=0

The zeros of 8x^2-4x+5=0 are x= 1/4+i 3/4 ,x= 1/4-i 3/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

8 x^{2} - 4 x + 5 = 0

x = \frac{1}{4} + i \frac{3}{4}, x = \frac{1}{4} - i \frac{3}{4}

Solution steps

8 x^{2} - 4 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 8 \cdot 5}{2 \cdot 8}

Simplify

(- 4)^{2} - 4 \cdot 8 \cdot 5 : 12 i

x_{1, 2} = \frac{- ( - 4 ) \pm 12 i}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 12 i}{2 \cdot 8}, x_{2} = \frac{- ( - 4 ) - 12 i}{2 \cdot 8}

x = \frac{- ( - 4 ) + 12 i}{2 \cdot 8} : \frac{1}{4} + i \frac{3}{4}

x = \frac{- ( - 4 ) - 12 i}{2 \cdot 8} : \frac{1}{4} - i \frac{3}{4}

The solutions to the quadratic equation are:

x = \frac{1}{4} + i \frac{3}{4}, x = \frac{1}{4} - i \frac{3}{4}

\frac{ln ( 4 )}{4}

\frac{- 2}{x} \leq x

- x^{2} - 3 x + 5 = 0

\frac{4}{5} (x - 2) \leq \frac{1}{3} (x - 6)

- 3 t = 18

What are the solutions to the equation 8x^2-4x+5=0 ?
The solutions to the equation 8x^2-4x+5=0 are x= 1/4+i 3/4 ,x= 1/4-i 3/4
Find the zeros of 8x^2-4x+5=0
The zeros of 8x^2-4x+5=0 are x= 1/4+i 3/4 ,x= 1/4-i 3/4