Question

1)Revolva:

x^{2}-8x+14=0

Answer 1

$x^{2} -8x+14=0$
Δ = b² - 4 * a * b
Δ = (-8)² - 4 * 1 * 14
Δ = 64 - 56
Δ = 8
√Δ = √8

$x'= \frac{-(-8) + \sqrt{8} }{2*1} = \frac{8+ \sqrt{8} }{2} = \frac{8+2 \sqrt{2} }{2} \\ \\ x'= 4+ \sqrt{2} \\ \\ x"= \frac{8-2 \sqrt{2} }{2} =4- \sqrt{2}$