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