What are the solutions to the equation x^2=(x-2)(x-2+x-1) ?

The solutions to the equation x^2=(x-2)(x-2+x-1) are x=6,x=1

Find the zeros of x^2=(x-2)(x-2+x-1)

The zeros of x^2=(x-2)(x-2+x-1) are x=6,x=1

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

x = 6, x = 1

Solution steps

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

Expand

(x - 2) (x - 2 + x - 1) : 2 x^{2} - 7 x + 6

x^{2} = 2 x^{2} - 7 x + 6

Switch sides

2 x^{2} - 7 x + 6 = x^{2}

Move

x^{2}

to the left side

x^{2} - 7 x + 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot 6}{2 \cdot 1}

(- 7)^{2} - 4 \cdot 1 \cdot 6 = 5

x_{1, 2} = \frac{- ( - 7 ) \pm 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 5}{2 \cdot 1}, x_{2} = \frac{- ( - 7 ) - 5}{2 \cdot 1}

x = \frac{- ( - 7 ) + 5}{2 \cdot 1} : 6

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

The solutions to the quadratic equation are:

x = 6, x = 1

3 x^{2} - 2 x - 21 = 0

r, s = 4 π r^{2}

9 (x^{2} - 9) - 2 = 17

x^{2} - 12 x + 36 = 44

5 x^{2} + 9 x + 4

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