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
