What are the solutions to the equation (6-x)(2-x)+5=0 ?

The solutions to the equation (6-x)(2-x)+5=0 are x=4+i,x=4-i

Find the zeros of (6-x)(2-x)+5=0

The zeros of (6-x)(2-x)+5=0 are x=4+i,x=4-i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(6 - x) (2 - x) + 5 = 0

x = 4 + i, x = 4 - i

Solution steps

(6 - x) (2 - x) + 5 = 0

Expand

(6 - x) (2 - x) + 5 : x^{2} - 8 x + 17

x^{2} - 8 x + 17 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot 17}{2 \cdot 1}

Simplify

(- 8)^{2} - 4 \cdot 1 \cdot 17 : 2 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 4 + i, x = 4 - i

6 x^{2} + 6 x - 13 = 0

2 x (x - 6) = 3

x^{2} - 4 x - 1 = 0

12 x^{2} - 4 = 13 x

(- x + 5)^{2} - 45 = 0

What are the solutions to the equation (6-x)(2-x)+5=0 ?
The solutions to the equation (6-x)(2-x)+5=0 are x=4+i,x=4-i
Find the zeros of (6-x)(2-x)+5=0
The zeros of (6-x)(2-x)+5=0 are x=4+i,x=4-i