[tex]I metoda\\
a) (x-1)(x+3)=0\\ \left \{ {{x-1=0} \atop {x+3=0}} \right. \ \textless \ =\ \textgreater \ \left \{ {{x=1} \atop {x=-3}} \right. \\
II metoda\\
(x-1)(x+3)=0\\ x^{2} +3x-x-3=0\\ x^{2} +2x-3=0\\
\Delta= b^{2}-4*a*c=2^2-4*1*(-3)=4+12=16\\
x_1= \frac{-b-\sqrt{\Delta}}{2*a}= \frac{-2-4}{2}= \frac{-6}{2}=-3\\
x_2= \frac{-b+\sqrt{\Delta}}{2*a}= \frac{-2+4}{2}= \frac{2}{2}=1[/tex]
Ambele metode sunt corecte. Sa iai in forma de totalitate dar nu sistem in prima metoda. Simbolul totalitatii aici nu e din pacate.
