What are the solutions to the equation (x-1)^2-13(x-1)+36=0 ?

The solutions to the equation (x-1)^2-13(x-1)+36=0 are x=10,x=5

Find the zeros of (x-1)^2-13(x-1)+36=0

The zeros of (x-1)^2-13(x-1)+36=0 are x=10,x=5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x - 1)^{2} - 13 (x - 1) + 36 = 0

x = 10, x = 5

Solution steps

(x - 1)^{2} - 13 (x - 1) + 36 = 0

Expand

(x - 1)^{2} - 13 (x - 1) + 36 : x^{2} - 15 x + 50

x^{2} - 15 x + 50 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 15 ) \pm ( - 15 ) ^{2} - 4 \cdot 1 \cdot 50}{2 \cdot 1}

(- 15)^{2} - 4 \cdot 1 \cdot 50 = 5

x_{1, 2} = \frac{- ( - 15 ) \pm 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 15 ) + 5}{2 \cdot 1}, x_{2} = \frac{- ( - 15 ) - 5}{2 \cdot 1}

x = \frac{- ( - 15 ) + 5}{2 \cdot 1} : 10

x = \frac{- ( - 15 ) - 5}{2 \cdot 1} : 5

The solutions to the quadratic equation are:

x = 10, x = 5

x^{4} - 14 x^{3} + 49 x^{2}

6 (x - 2)^{2} + 24 (x - 2) - 1 = 0

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

8 x - 21 = 2 x - 9

What are the solutions to the equation (x-1)^2-13(x-1)+36=0 ?
The solutions to the equation (x-1)^2-13(x-1)+36=0 are x=10,x=5
Find the zeros of (x-1)^2-13(x-1)+36=0
The zeros of (x-1)^2-13(x-1)+36=0 are x=10,x=5