What are the solutions to the equation (x-3)^2+2(x-3)-8=0 ?

The solutions to the equation (x-3)^2+2(x-3)-8=0 are x=5,x=-1

Find the zeros of (x-3)^2+2(x-3)-8=0

The zeros of (x-3)^2+2(x-3)-8=0 are x=5,x=-1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x - 3)^{2} + 2 (x - 3) - 8 = 0

x = 5, x = - 1

Solution steps

(x - 3)^{2} + 2 (x - 3) - 8 = 0

Expand

(x - 3)^{2} + 2 (x - 3) - 8 : x^{2} - 4 x - 5

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

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot (- 5) = 6

x_{1, 2} = \frac{- ( - 4 ) \pm 6}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 6}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 6}{2 \cdot 1}

x = \frac{- ( - 4 ) + 6}{2 \cdot 1} : 5

x = \frac{- ( - 4 ) - 6}{2 \cdot 1} : - 1

The solutions to the quadratic equation are:

x = 5, x = - 1

\frac{12}{- 3}

1 < 3 - 2 x < 7

\frac{( 2 x + 3 )}{4} - \frac{( 3 - x )}{5} = \frac{( x + 2 )}{2}

\frac{x ^{- 7}}{x ^{- 8}}

ln (\frac{1}{e ^{6}})

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