What is the answer to a=(x-2)^2+(x+3)^2-2(x+1)(x-1) ?

The answer to a=(x-2)^2+(x+3)^2-2(x+1)(x-1) is x=(a-15)/2

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a=(x-2)^2+(x+3)^2-2(x+1)(x-1)

a = (x - 2)^{2} + (x + 3)^{2} - 2 (x + 1) (x - 1)

x = \frac{a - 15}{2}

Solution steps

a = (x - 2)^{2} + (x + 3)^{2} - 2 (x + 1) (x - 1)

Switch sides

(x - 2)^{2} + (x + 3)^{2} - 2 (x + 1) (x - 1) = a

Expand

(x - 2)^{2} + (x + 3)^{2} - 2 (x + 1) (x - 1) : 2 x + 15

2 x + 15 = a

Move

15

to the right side

2 x = a - 15

Divide both sides by

2

x = \frac{a - 15}{2}

2 x + 3 = \frac{x}{2} - 1

x + 1 = (2^{8} mod 7)

2 (x - 4) = x - 7

2 (5 x - 4) = 6 + (3 x - 4)

x, 32 x = \frac{y}{2}

What is the answer to a=(x-2)^2+(x+3)^2-2(x+1)(x-1) ?
The answer to a=(x-2)^2+(x+3)^2-2(x+1)(x-1) is x=(a-15)/2