Question

$x^{2} + 4x - 5 =0$  como resolver com Bhaskara?

Answer 1

b²- 4.a.c
4² - 4.1.(-5)
16 + 20
Δ = 36
[-(b) +/-√Δ] / 2a
(-4 +/-√36) / 2.1
x1 = -4 + 6 / 2
x1 = 2/2 = 1
x2 = -4 - 6 / 2
x2 = -10 / 2
x2 = -5
Logo, x1 = 1 e x2 = -5