solve for (x-4)^2=13
\((x-4)^2=13\)
First take the square root of both sides.
\(x-4= \sqrt{13}\), \(- \sqrt{13}\). There are two solutions because a square root comes with a positive and negative solution.
Now move the 4 over by adding 4 on both sides.
\(x= \sqrt{13}+4\), \(- \sqrt{13} +4\)
(x - 4)² = 13
(x - 4)(x - 4) = 13
x(x - 4) - 4(x - 4) = 13
x(x) - x(4) - 4(x) + 4(4) = 13
x² - 4x - 4x + 16 = 13
x² - 8x + 16 = 13
- 13 - 13
x² - 8x + 3 = 0
x = -(-8) ± √((-8)² - 4(1)(3))
2(1)
x = 8 ± √(64 - 12)
2
x = 8 ± √(52)
2
x = 8 ± 2√(13)
2
x = 4 ± √(13)
x = 4 + √(13) or x = 4 - √(13)