Solve for $x$, where $x>0$ and $5x^2+9x-18=0$. Express your answer as a simplified common fraction.
I notice that this equation is a quadratic equation, which means it can be solved by factoring, using the quadratic formula, or completing the square. I also notice that the leading coefficient, 5, is not 1, so factoring might be more difficult than usual. I decide to try the quadratic formula, which says that if $ax^2+bx+c=0$, then $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. In this case, $a=5$, $b=9$, and $c=-18$, so I plug these values into the formula and simplify. I get $x=\frac{-9\pm\sqrt{81+360}}{10}$. The square root of 441 is 21, so I can simplify further and get $x=\frac{-9\pm 21}{10}$. This gives me two possible values for $x$: $x=\frac{12}{10}$ or $x=\frac{-30}{10}$. However, the problem says that $x>0$, so I can eliminate the negative value and only keep the positive one. I can reduce the fraction $\frac{12}{10}$ by dividing both the numerator and the denominator by 2, and get $x=\frac{6}{5}$. This is my final answer. 