What are the solutions to the equation (x-4)(x-3)+2=x-1 ?

The solutions to the equation (x-4)(x-3)+2=x-1 are x=5,x=3

Find the zeros of (x-4)(x-3)+2=x-1

The zeros of (x-4)(x-3)+2=x-1 are x=5,x=3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x - 4) (x - 3) + 2 = x - 1

x = 5, x = 3

Solution steps

(x - 4) (x - 3) + 2 = x - 1

Expand

(x - 4) (x - 3) + 2 : x^{2} - 7 x + 14

x^{2} - 7 x + 14 = x - 1

Move

1

to the left side

x^{2} - 7 x + 15 = x

Move

x

to the left side

x^{2} - 8 x + 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot 15}{2 \cdot 1}

(- 8)^{2} - 4 \cdot 1 \cdot 15 = 2

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

Separate the solutions

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

x = \frac{- ( - 8 ) + 2}{2 \cdot 1} : 5

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

The solutions to the quadratic equation are:

x = 5, x = 3

\frac{x ^{2} - 1}{6} - \frac{x - 1}{4} = 3

2 (x - 8)^{2} - 35 = 7

x^{2} - 5 x + 6 = 0

w, 30 w = 9 w^{2} + 25

d x^{2}

What are the solutions to the equation (x-4)(x-3)+2=x-1 ?
The solutions to the equation (x-4)(x-3)+2=x-1 are x=5,x=3
Find the zeros of (x-4)(x-3)+2=x-1
The zeros of (x-4)(x-3)+2=x-1 are x=5,x=3