Question

Найдите корни квадратного уравнения 2x^2-5x-3=0

Answer 1

D=b^2-4ac=25+24=49
x1=(-b+$sqrt{D}$)/2a=(5+7)/4=3
x2=-b-$sqrt{D}$)/2a=(5-7)/4=-1/2

Answer 2

2*(x²-5*x/2-3/2)=2*[(x-5/4)²-49/16]=0, (x-5/4)²-49/16=0, (x-5/4)²=49/16,
x-5/4=7/4  ⇒ x=3
x-5/4=-7/4 ⇒x=-1/2
Ответ: x1=3, x2=-1/2.