Question

4x^{2} + x + 12 = 0 (me ajudem)

Answer 1

4x^2 + x + 12 = 0

Δ = b - 4.a.c 
Δ = 0^2 - 4 . 1 . -100 
Δ = 0^2 - 4. 1 . -100 
Δ = 400

Há 2 raízes reais.

x = (-b +- √Δ)/2a

x' = (-0 + √400)/2.1   
x'' = (-0 - √400)/2.1

x' = 20 / 2   
x'' = -20 / 2

x' = 10   
x'' = -10