What are the solutions to the equation 5=(x-2)(x-6) ?

The solutions to the equation 5=(x-2)(x-6) are x=7,x=1

Find the zeros of 5=(x-2)(x-6)

The zeros of 5=(x-2)(x-6) are x=7,x=1

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5 = (x - 2) (x - 6)

x = 7, x = 1

Solution steps

5 = (x - 2) (x - 6)

Expand

(x - 2) (x - 6) : x^{2} - 8 x + 12

5 = x^{2} - 8 x + 12

Switch sides

x^{2} - 8 x + 12 = 5

Move

5

to the left side

x^{2} - 8 x + 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot 7}{2 \cdot 1}

(- 8)^{2} - 4 \cdot 1 \cdot 7 = 6

x_{1, 2} = \frac{- ( - 8 ) \pm 6}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 6}{2 \cdot 1}, x_{2} = \frac{- ( - 8 ) - 6}{2 \cdot 1}

x = \frac{- ( - 8 ) + 6}{2 \cdot 1} : 7

x = \frac{- ( - 8 ) - 6}{2 \cdot 1} : 1

The solutions to the quadratic equation are:

x = 7, x = 1

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

x \cdot (1 + x) = 1

6 x^{2} - 9 x - 3 = 0

3 n - n^{2} = 0

20 = 2 + 35 t - 16 t^{2}

What are the solutions to the equation 5=(x-2)(x-6) ?
The solutions to the equation 5=(x-2)(x-6) are x=7,x=1
Find the zeros of 5=(x-2)(x-6)
The zeros of 5=(x-2)(x-6) are x=7,x=1