What is the answer to (8+x)^2-x(7x-1)=(x-8)^2-8x-7(x+3)(x-3) ?

The answer to (8+x)^2-x(7x-1)=(x-8)^2-8x-7(x+3)(x-3) is x= 63/41

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(8+x)^2-x(7x-1)=(x-8)^2-8x-7(x+3)(x-3)

(8 + x)^{2} - x (7 x - 1) = (x - 8)^{2} - 8 x - 7 (x + 3) (x - 3)

x = \frac{63}{41}

Decimal Notation

x = 1.53658 \dots

Solution steps

(8 + x)^{2} - x (7 x - 1) = (x - 8)^{2} - 8 x - 7 (x + 3) (x - 3)

Expand

(8 + x)^{2} - x (7 x - 1) : - 6 x^{2} + 17 x + 64

Expand

(x - 8)^{2} - 8 x - 7 (x + 3) (x - 3) : - 6 x^{2} - 24 x + 127

- 6 x^{2} + 17 x + 64 = - 6 x^{2} - 24 x + 127

Move

64

to the right side

- 6 x^{2} + 17 x = - 6 x^{2} - 24 x + 63

Subtract

- 6 x^{2} - 24 x

from both sides

- 6 x^{2} + 17 x - (- 6 x^{2} - 24 x) = - 6 x^{2} - 24 x + 63 - (- 6 x^{2} - 24 x)

Simplify

41 x = 63

Divide both sides by

41

x = \frac{63}{41}

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

\frac{6}{5} x - \frac{7}{12} = 3 x + \frac{2}{5}

\frac{2}{5} a + 3 = 23

x, 8 x - 6 y = 0

\frac{5 x + 1}{3} - \frac{( x + 1 ) ^{2}}{2} = 1 - \frac{x ^{2}}{2} + \frac{1}{6}

What is the answer to (8+x)^2-x(7x-1)=(x-8)^2-8x-7(x+3)(x-3) ?
The answer to (8+x)^2-x(7x-1)=(x-8)^2-8x-7(x+3)(x-3) is x= 63/41
Solve for x: (8+x)^2-x(7x-1)=(x-8)^2-8x-7(x+3)(x-3)
The solution is x= 63/41