Question

4x^2-19x+12=0 help pls

Answer 1

a=4, b = -19, c=12
D = (-19)^2 - 4*4*12 = 169
x1 = (-(-19) + sqrt(169))/2*4 = (19+13)/8 = 4
x2 = (-(-19) - sqrt(169))/2*4 = (19-13)/8 = 0.75

Ответ: 4 и 0ю75