What are the solutions to the equation (x-4)^2-5(x-4)=24 ?

The solutions to the equation (x-4)^2-5(x-4)=24 are x=12,x=1

Find the zeros of (x-4)^2-5(x-4)=24

The zeros of (x-4)^2-5(x-4)=24 are x=12,x=1

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(x - 4)^{2} - 5 (x - 4) = 24

x = 12, x = 1

Solution steps

(x - 4)^{2} - 5 (x - 4) = 24

Expand

(x - 4)^{2} - 5 (x - 4) : x^{2} - 13 x + 36

x^{2} - 13 x + 36 = 24

Move

24

to the left side

x^{2} - 13 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

(- 13)^{2} - 4 \cdot 1 \cdot 12 = 11

x_{1, 2} = \frac{- ( - 13 ) \pm 11}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 13 ) + 11}{2 \cdot 1}, x_{2} = \frac{- ( - 13 ) - 11}{2 \cdot 1}

x = \frac{- ( - 13 ) + 11}{2 \cdot 1} : 12

x = \frac{- ( - 13 ) - 11}{2 \cdot 1} : 1

The solutions to the quadratic equation are:

x = 12, x = 1

a^{2} - 2 a - 8 = 0

- 20 = 10 t - \frac{9.81 t ^{2}}{2}

30 + 19 x - 4 x^{2} = 0

x, b x = a x^{2} + c

5 x - 6 = - 2 x^{2}

What are the solutions to the equation (x-4)^2-5(x-4)=24 ?
The solutions to the equation (x-4)^2-5(x-4)=24 are x=12,x=1
Find the zeros of (x-4)^2-5(x-4)=24
The zeros of (x-4)^2-5(x-4)=24 are x=12,x=1