What are the solutions to the equation r^2-3r+4=0 ?

The solutions to the equation r^2-3r+4=0 are r= 3/2+i(sqrt(7))/2 ,r= 3/2-i(sqrt(7))/2

Find the zeros of r^2-3r+4=0

The zeros of r^2-3r+4=0 are r= 3/2+i(sqrt(7))/2 ,r= 3/2-i(sqrt(7))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

r^{2} - 3 r + 4 = 0

r = \frac{3}{2} + i \frac{7}{2}, r = \frac{3}{2} - i \frac{7}{2}

Solution steps

r^{2} - 3 r + 4 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot 4}{2 \cdot 1}

Simplify

(- 3)^{2} - 4 \cdot 1 \cdot 4 : 7 i

r_{1, 2} = \frac{- ( - 3 ) \pm 7 i}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- ( - 3 ) + 7 i}{2 \cdot 1}, r_{2} = \frac{- ( - 3 ) - 7 i}{2 \cdot 1}

r = \frac{- ( - 3 ) + 7 i}{2 \cdot 1} : \frac{3}{2} + i \frac{7}{2}

r = \frac{- ( - 3 ) - 7 i}{2 \cdot 1} : \frac{3}{2} - i \frac{7}{2}

The solutions to the quadratic equation are:

r = \frac{3}{2} + i \frac{7}{2}, r = \frac{3}{2} - i \frac{7}{2}

z^{2} + 4 z = 2

- n + 2 - n < - \frac{2}{3} (9 n - 63)

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

500 x^{3} + 108 y^{18}

- 3 (5 - 9 v) = 25 + 7 v

What are the solutions to the equation r^2-3r+4=0 ?
The solutions to the equation r^2-3r+4=0 are r= 3/2+i(sqrt(7))/2 ,r= 3/2-i(sqrt(7))/2
Find the zeros of r^2-3r+4=0
The zeros of r^2-3r+4=0 are r= 3/2+i(sqrt(7))/2 ,r= 3/2-i(sqrt(7))/2