What are the solutions to the equation (x-6)(x-3)=4 ?

The solutions to the equation (x-6)(x-3)=4 are x=7,x=2

Find the zeros of (x-6)(x-3)=4

The zeros of (x-6)(x-3)=4 are x=7,x=2

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(x - 6) (x - 3) = 4

x = 7, x = 2

Solution steps

(x - 6) (x - 3) = 4

Expand

(x - 6) (x - 3) : x^{2} - 9 x + 18

x^{2} - 9 x + 18 = 4

Move

4

to the left side

x^{2} - 9 x + 14 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 1 \cdot 14}{2 \cdot 1}

(- 9)^{2} - 4 \cdot 1 \cdot 14 = 5

x_{1, 2} = \frac{- ( - 9 ) \pm 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 5}{2 \cdot 1}, x_{2} = \frac{- ( - 9 ) - 5}{2 \cdot 1}

x = \frac{- ( - 9 ) + 5}{2 \cdot 1} : 7

x = \frac{- ( - 9 ) - 5}{2 \cdot 1} : 2

The solutions to the quadratic equation are:

x = 7, x = 2

(6 - x)^{2} - 1 = 0

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

x, t x^{2} - 12 x + 9 = 0

(2 x + 1)^{2} = 1

4 (5 x - 2)^{2} - 9 = 4

What are the solutions to the equation (x-6)(x-3)=4 ?
The solutions to the equation (x-6)(x-3)=4 are x=7,x=2
Find the zeros of (x-6)(x-3)=4
The zeros of (x-6)(x-3)=4 are x=7,x=2