Question

Resolver a equação em R
2(x+3)^2-5(x+3)+2=0

Answer 1

Explicação passo-a-passo:

2.(x+3)² - 5.(x+3) +2=0

2.(x+3).(x+3) - 5x -15 +2=0

2.(x²+3x +3x +9) -5x - 13=0

2.( x²+6x +9) -5x -13=0

2x² +12x +18 -5x -13=0

2x² +7x + 5=0

∆= 7² -4.2.5

∆ = 49 - 40

∆= 9

x= -7 ±√9/2.2

x= -7 ±3/4

x1= -7-3/4

x1= -10/4

x1= - 5/2

x2= -7+3/4

x2= -4/4

x2= -1

S={ - 1, -5/2}