What are the solutions to the equation 2x^2-7x+8=0 ?

The solutions to the equation 2x^2-7x+8=0 are x= 7/4+i(sqrt(15))/4 ,x= 7/4-i(sqrt(15))/4

Find the zeros of 2x^2-7x+8=0

The zeros of 2x^2-7x+8=0 are x= 7/4+i(sqrt(15))/4 ,x= 7/4-i(sqrt(15))/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

x = \frac{7}{4} + i \frac{15}{4}, x = \frac{7}{4} - i \frac{15}{4}

Solution steps

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 \cdot 8}{2 \cdot 2}

Simplify

(- 7)^{2} - 4 \cdot 2 \cdot 8 : 15 i

x_{1, 2} = \frac{- ( - 7 ) \pm 15 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 15 i}{2 \cdot 2}, x_{2} = \frac{- ( - 7 ) - 15 i}{2 \cdot 2}

x = \frac{- ( - 7 ) + 15 i}{2 \cdot 2} : \frac{7}{4} + i \frac{15}{4}

x = \frac{- ( - 7 ) - 15 i}{2 \cdot 2} : \frac{7}{4} - i \frac{15}{4}

The solutions to the quadratic equation are:

x = \frac{7}{4} + i \frac{15}{4}, x = \frac{7}{4} - i \frac{15}{4}

\frac{1}{4} \leq \frac{6 - x}{12} \leq - \frac{1}{6}

\frac{1}{3} x + 2 < \frac{1}{6} x - 1

\frac{p}{4} + 8 = 7

12 + 8

\frac{3}{z - 2} \geq 4

What are the solutions to the equation 2x^2-7x+8=0 ?
The solutions to the equation 2x^2-7x+8=0 are x= 7/4+i(sqrt(15))/4 ,x= 7/4-i(sqrt(15))/4
Find the zeros of 2x^2-7x+8=0
The zeros of 2x^2-7x+8=0 are x= 7/4+i(sqrt(15))/4 ,x= 7/4-i(sqrt(15))/4