Question

encontre
${x}^{1} {x}^{2} \\ - 10x + 22 = 0 a$

Answer 1

x^2 - 10x + 22 = 0

a = 1; b = - 10; c = 22

∆ = b^2-4ac

∆ = (-10)^2 - 4.1.22

∆ = 100 - 88

∆ = 12

√∆ = √12 = √4.√3 = 2√3

x = [ - b +/- √∆]/2a

x = [ -(-10) +/- √12]/2.1

x = [10 +/- 2√3]/2

x = [ 10/2 +/- 2√3/2]

x' = 5 + √3

x" = 5 - √3