Explanation:
let's assume you really meant
2(7 + 1)7 = (2x + 2)²
2×8×7 = (2x + 2)(2x + 2)
112 = 4x² + 4x + 4x + 4 = 4x² + 8x + 4
let's divide everything by 4 to get a simpler equation :
28 = x² + 2x + 1
x² + 2x - 27 = 0
the general solution to such a quadratic equation
ax² + bx + c = 0
is
x = (-b ± sqrt(b² - 4ac))/(2a)
so, in our case
x = (-2 ± sqrt(2² - 4×1×-27))/(2×1) =
= (-2 ± sqrt(4 + 108))/2 = (-2 ± sqrt(112))/2 =
= (-2 ± sqrt(16×7))/2 = (-2 ± 4×sqrt(7))/2 =
= -1 ± 2×sqrt(7)
x1 = -1 + 2×sqrt(7) = 4.291502622...
x2 = -1 - 2×sqrt(7) = -6.291502622...
so, that equation is only an identity for the 2 x-values we found.
for any other x-value it is not.
that is why I suspect you skipped or changed something in the original equation.
could it be it was
(2(x + 1))² = (2x + 2)²
then we would have
4(x + 1)² = 4(x² + 2x + 1) = 4x² + 8x + 4
4x² + 8x + 4 = 4x² + 8x + 4
the equality is given because
2(x + 1) = 2x + 2
please let me know, if it was something else. but given the wording in the question I doubt that your equation really was what you wrote, and what I solved in the first step.