What are the solutions to the equation (x-4)^2-13(x-4)+36=0 ?

The solutions to the equation (x-4)^2-13(x-4)+36=0 are x=13,x=8

Find the zeros of (x-4)^2-13(x-4)+36=0

The zeros of (x-4)^2-13(x-4)+36=0 are x=13,x=8

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(x - 4)^{2} - 13 (x - 4) + 36 = 0

x = 13, x = 8

Solution steps

(x - 4)^{2} - 13 (x - 4) + 36 = 0

Expand

(x - 4)^{2} - 13 (x - 4) + 36 : x^{2} - 21 x + 104

x^{2} - 21 x + 104 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 21 ) \pm ( - 21 ) ^{2} - 4 \cdot 1 \cdot 104}{2 \cdot 1}

(- 21)^{2} - 4 \cdot 1 \cdot 104 = 5

x_{1, 2} = \frac{- ( - 21 ) \pm 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 21 ) + 5}{2 \cdot 1}, x_{2} = \frac{- ( - 21 ) - 5}{2 \cdot 1}

x = \frac{- ( - 21 ) + 5}{2 \cdot 1} : 13

x = \frac{- ( - 21 ) - 5}{2 \cdot 1} : 8

The solutions to the quadratic equation are:

x = 13, x = 8

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

x, y = (x - y)^{2}

4 - 25 x^{2} = 0

x^{2} - 2 x + 10

9 m^{2} + 24 m + 16 = 0

What are the solutions to the equation (x-4)^2-13(x-4)+36=0 ?
The solutions to the equation (x-4)^2-13(x-4)+36=0 are x=13,x=8
Find the zeros of (x-4)^2-13(x-4)+36=0
The zeros of (x-4)^2-13(x-4)+36=0 are x=13,x=8