y²-(y1+y2)y+y1y2=0
y²-(x1+x2-2)y+(x1x2-x1-x2+1)=0
dar
x1+x2=-(-2)/1=2
si x1x2=-4
atunci
y²-(2-2)y+(-4-2+1)=0
y²-5=0
y²-(y1+y2)y+y1y2=0
y²-((x1²+x2²)/x1x2) *y +1=0
dar x1²+x2²=(x1+x2)²-2x1x2=4-2*(-4)=4+8=12
deci
y²-(12/(-4) )*y+1=0
y²+3y+1=0
fac o verificare la prima pt ca a rezultat
x1.2= (2+-√(4+16))/2=(2+-2√5)/2=(1+-√5)
y1,2=+-√5 , e bine