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

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

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

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

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

x = 6

Solution steps

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

Expand

(x - 8) (x - 4) : 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

x, 2 x^{2} + 4 y^{2} + 8 x - 8 y + 12 = 0

(2 x - 1) (6 x + 8) = 4 (x^{2} - 5)

\frac{2 x ^{2} - 6}{2} = \frac{15 x - 9}{6} - \frac{3 x + 14}{8}

100 = 2 a^{2}

2 x^{2} - 7 x = 0

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