So 2 real solutions just write 2 factors and expand (x-3)(x-4)=0 x^2-7x+12=0 discriminant is when you have b^2-4ac so subsitute ax^2+bx+c=0 a=1 b=-7 c=12 -7^2-4(1)(12) 49-48=1 discriminant=1
one real solution would be (x-1)(x-1)=0 x^2-2x+1=0 discriminant=-2^2-4(1)(1)=4-4=0 discriminant=0
no real solutions has the discriminant less than zero let's say that b=1 and c=20 and a=1 basically, we need a set of numbers where b^2 is less than 4acso
1^2-4(1)(20)=1-80=-79 then we subsitute x^2+x+20=0
