What are the solutions to the equation x(x-4)+(x-2)=2 ?

The solutions to the equation x(x-4)+(x-2)=2 are x=4,x=-1

Find the zeros of x(x-4)+(x-2)=2

The zeros of x(x-4)+(x-2)=2 are x=4,x=-1

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x (x - 4) + (x - 2) = 2

x = 4, x = - 1

Solution steps

x (x - 4) + (x - 2) = 2

Expand

x (x - 4) + (x - 2) : x^{2} - 3 x - 2

x^{2} - 3 x - 2 = 2

Move

2

to the left side

x^{2} - 3 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 4 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 4) = 5

x_{1, 2} = \frac{- ( - 3 ) \pm 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 5}{2 \cdot 1}, x_{2} = \frac{- ( - 3 ) - 5}{2 \cdot 1}

x = \frac{- ( - 3 ) + 5}{2 \cdot 1} : 4

x = \frac{- ( - 3 ) - 5}{2 \cdot 1} : - 1

The solutions to the quadratic equation are:

x = 4, x = - 1

x + 2 > 10

25 x^{2} + 10 x + 1 = 0

16 \div 4 (4)

b^{2} + 49 - (b - 7)^{2}

What are the solutions to the equation x(x-4)+(x-2)=2 ?
The solutions to the equation x(x-4)+(x-2)=2 are x=4,x=-1
Find the zeros of x(x-4)+(x-2)=2
The zeros of x(x-4)+(x-2)=2 are x=4,x=-1