What are the solutions to the equation (x-5)^2-4(x-5)-21=0 ?

The solutions to the equation (x-5)^2-4(x-5)-21=0 are x=12,x=2

Find the zeros of (x-5)^2-4(x-5)-21=0

The zeros of (x-5)^2-4(x-5)-21=0 are x=12,x=2

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(x - 5)^{2} - 4 (x - 5) - 21 = 0

x = 12, x = 2

Solution steps

(x - 5)^{2} - 4 (x - 5) - 21 = 0

Expand

(x - 5)^{2} - 4 (x - 5) - 21 : x^{2} - 14 x + 24

x^{2} - 14 x + 24 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 1 \cdot 24}{2 \cdot 1}

(- 14)^{2} - 4 \cdot 1 \cdot 24 = 10

x_{1, 2} = \frac{- ( - 14 ) \pm 10}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 14 ) + 10}{2 \cdot 1}, x_{2} = \frac{- ( - 14 ) - 10}{2 \cdot 1}

x = \frac{- ( - 14 ) + 10}{2 \cdot 1} : 12

x = \frac{- ( - 14 ) - 10}{2 \cdot 1} : 2

The solutions to the quadratic equation are:

x = 12, x = 2

w^{2} - 4 w + 4 = 0

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

lo g_{10} (5)

lo g_{0.5} (1)

\frac{2}{- 10}

What are the solutions to the equation (x-5)^2-4(x-5)-21=0 ?
The solutions to the equation (x-5)^2-4(x-5)-21=0 are x=12,x=2
Find the zeros of (x-5)^2-4(x-5)-21=0
The zeros of (x-5)^2-4(x-5)-21=0 are x=12,x=2