What are the solutions to the equation (x-4)^2-10(x-4)+24=0 ?

The solutions to the equation (x-4)^2-10(x-4)+24=0 are x=10,x=8

Find the zeros of (x-4)^2-10(x-4)+24=0

The zeros of (x-4)^2-10(x-4)+24=0 are x=10,x=8

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(x - 4)^{2} - 10 (x - 4) + 24 = 0

x = 10, x = 8

Solution steps

(x - 4)^{2} - 10 (x - 4) + 24 = 0

Expand

(x - 4)^{2} - 10 (x - 4) + 24 : x^{2} - 18 x + 80

x^{2} - 18 x + 80 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 1 \cdot 80}{2 \cdot 1}

(- 18)^{2} - 4 \cdot 1 \cdot 80 = 2

x_{1, 2} = \frac{- ( - 18 ) \pm 2}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 18 ) + 2}{2 \cdot 1}, x_{2} = \frac{- ( - 18 ) - 2}{2 \cdot 1}

x = \frac{- ( - 18 ) + 2}{2 \cdot 1} : 10

x = \frac{- ( - 18 ) - 2}{2 \cdot 1} : 8

The solutions to the quadratic equation are:

x = 10, x = 8

\frac{a}{a ^{2}}

(13 m^{4} + 2) + (m^{4} n^{2} + 2 - 2 m^{4}) - (- 13 m^{2} n^{3} + 5 m^{4})

22500 - 740 x = 15200 + 1500 x

(x + 8)^{2} - 4

\frac{x}{4} = 12

What are the solutions to the equation (x-4)^2-10(x-4)+24=0 ?
The solutions to the equation (x-4)^2-10(x-4)+24=0 are x=10,x=8
Find the zeros of (x-4)^2-10(x-4)+24=0
The zeros of (x-4)^2-10(x-4)+24=0 are x=10,x=8