What are the solutions to the equation (x-4)(x-8)=-4 ?

The solutions to the equation (x-4)(x-8)=-4 are x=6

Find the zeros of (x-4)(x-8)=-4

The zeros of (x-4)(x-8)=-4 are x=6

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(x - 4) (x - 8) = - 4

x = 6

Solution steps

(x - 4) (x - 8) = - 4

Expand

(x - 4) (x - 8) : x^{2} - 12 x + 32

x^{2} - 12 x + 32 = - 4

Move

4

to the left side

x^{2} - 12 x + 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 36}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot 36 = 0

x_{1, 2} = \frac{- ( - 12 ) \pm 0}{2 \cdot 1}

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

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

x = 6

The solution to the quadratic equation is:

x = 6

- 2 x^{2} - 4 x + 13 = 0

x^{2} - 4 x = 5 x

2 x^{2} + 9 x + j = 0

h^{2} + 3 h - 1 = 0

n (n - 3) = 4 n

What are the solutions to the equation (x-4)(x-8)=-4 ?
The solutions to the equation (x-4)(x-8)=-4 are x=6
Find the zeros of (x-4)(x-8)=-4
The zeros of (x-4)(x-8)=-4 are x=6