Question

4x²-19x÷2 - 2,5=0
Помогите плиииз

Answer 1

Ответ:

4x^2-19x-5 = 0

D= (-19)^2 - 4*4*(-5) = 361+80 = 441

x1= (19+21) / (2*4) = 40/8 = 5

x2= (19-21) / (2*4) =- 2/8 = -1/4