For the equation x^2+3x+j=0 , find all the values of j such that the equation has two real number solutions. show your work.

Use quadrratic equation discriminant
for
ax²+bx+c=0

if b²-4ac is the discriminant

if it is greater than 0, then it has 2 real number solutions
if it is equal to 0 then it has 1 real number solution
if it is less than 0 then it has 0 real number solutions

so

1x²+3x+j=0
b²-4ac=3²-4(1)(j)
9-4j
to have it have a real number solution, it must be greater than 0
9-4j>0
9>4j
9/4>j

j<9/4 is the solution area