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

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

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

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

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

x = 10, x = 5

Solution steps

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

Expand

(x - 1)^{2} - 13 (x - 1) : x^{2} - 15 x + 14

x^{2} - 15 x + 14 = - 36

Move

36

to the left side

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

2 x^{2} - 7 x + 5 = 0

(x - 3) (1 - x) = 1

2 x + 7 \leq 12

(x + 9) (x - 4) < 0

7 (x - 5) - 2 x (x - 5)

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