Question

x^+5x+4=0 determine o valor de ∆=b^-4.a.c

Answer 1

∆=b^2-4.a.c
∆=5^2-4.1.4
∆=25 - 16
∆= 9

^^.

Answer 2

x² + 5x + 4 = 0
Δ = b² - 4ac
Δ = 5² - 4 * 1 * 4
Δ = 25 - 16
Δ = 9

x' = -5 + √9 / 2 * 1
x' = -5 + 3 / 2
x' = -2 / 2
x' = -1

x" = -5 - 3 / 2
x" = -8 / 2
x" = -4