2x^2-10x+5=2(x^2-5x+5/2)=2x2−10x+5=2(x2−5x+52)=
=2(x^2-2*x*5/2+(5/2)^2-(5/2)^2+5/2)==2(x2−2⋅x⋅52+(52)2−(52)2+52)=
2((x-5/2)^2-25/4+10/4)=2((x-5/2)^2-15/4)=2((x−52)2−254+104)=2((x−52)2−154)=
=2((x-5/2)^2-(sqrt15/2)^2)==2⎛⎝(x−52)2−(√152)2⎞⎠=
=2(x-5/2-sqrt15/2)(x-5/2+sqrt15/2)==2(x−52−√152)(x−52+√152)=
=2(x-(5+sqrt15)/2)(x-(5-sqrt15)/2)=2(x−5+√152)(x−5−√152)
General case:
P_2(x)=ax^2+bx+c=a(x^2+b/ax+c/a)=P2(x)=ax2+bx+c=a(x2+bax+ca)=
=a(x^2+2*x*b/(2a)+(b/(2a))^2-(b/(2a))^2+c/a)==a(x2+2⋅x⋅b2a+(b2a)2−(b2a)2+ca)=
=a((x+b/(2a))^2-(b^2-4ac)/(4a^2))==a((x+b2a)2−b2−4ac4a2)=
=a((x+b/(2a))^2-(sqrt(b^2-4ac)/(2a))^2)==a⎛⎝(x+b2a)2−(√b2−4ac2a)2⎞⎠=
=a(x+b/(2a)-sqrt(b^2-4ac)/(2a))(x+b/(2a)+sqrt(b^2-4ac)/(2a))==a(x+b2a−√b2−4ac2a)(x+b2a+√b2−4ac2a)=
=a(x-(-b+sqrt(b^2-4ac))/(2a))(x-(-b-sqrt(b^2-4ac))/(2a))=a(x−−b+√b2−4ac2a)(x−−b−√b2−4ac2a)
You can see that P_2(x)=0P2(x)=0 for
x-(-b+sqrt(b^2-4ac))/(2a)=0 => x=(-b+sqrt(b^2-4ac))/(2a)x−−b+√b2−4ac2a=0⇒x=−b+√b2−4ac2a
or
x+(-b+sqrt(b^2-4ac))/(2a)=0 => x=(-b-sqrt(b^2-4ac))/(2a)x+−b+√b2−4ac2a=0⇒x=−b−√b2−4ac2a
which is famous formula for solutions of the quadratic equation.
