What are the solutions to the equation (x-1)^2-4(x-1)-32=0 ?

The solutions to the equation (x-1)^2-4(x-1)-32=0 are x=9,x=-3

Find the zeros of (x-1)^2-4(x-1)-32=0

The zeros of (x-1)^2-4(x-1)-32=0 are x=9,x=-3

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(x - 1)^{2} - 4 (x - 1) - 32 = 0

x = 9, x = - 3

Solution steps

(x - 1)^{2} - 4 (x - 1) - 32 = 0

Expand

(x - 1)^{2} - 4 (x - 1) - 32 : x^{2} - 6 x - 27

x^{2} - 6 x - 27 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot ( - 27 )}{2 \cdot 1}

(- 6)^{2} - 4 \cdot 1 \cdot (- 27) = 12

x_{1, 2} = \frac{- ( - 6 ) \pm 12}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 12}{2 \cdot 1}, x_{2} = \frac{- ( - 6 ) - 12}{2 \cdot 1}

x = \frac{- ( - 6 ) + 12}{2 \cdot 1} : 9

x = \frac{- ( - 6 ) - 12}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

x = 9, x = - 3

x^{2} - 3 x - 28 = 0

81 \cdot 3

\frac{11}{2}

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

- 5 y^{2} - 8 y + 4

What are the solutions to the equation (x-1)^2-4(x-1)-32=0 ?
The solutions to the equation (x-1)^2-4(x-1)-32=0 are x=9,x=-3
Find the zeros of (x-1)^2-4(x-1)-32=0
The zeros of (x-1)^2-4(x-1)-32=0 are x=9,x=-3