Answer:
2 different solutions
Step-by-step explanation:
[tex]\text{The discriminant of a quadratic equation}\ ax^2+bx+c=0:\\\\\Delta=b^2-4ac\\\\\text{If}\ \Delta<0,\ \text{then the equation has no solution}\\\\\text{If}\ \Delta=0,\ \text{then the equation has one solution}\ x=\dfrac{-b}{2a}\\\\\text{If}\ \Delta>0,\ \text{then the wquation has two different solutions}\ x=\dfrac{-b\pm\sqrt\Delta}{2a}[/tex]
[tex]\text{We have the equation}\ x^2+3.01x+2.25=0\to a=1,\ b=3.01,\ c=2.25.\\\\\text{Substitute:}\\\\b^2-4ac=(3.01)^2+4(1)(2.25)=9.0601+10=19.0601>0[/tex]
