Quadratic equation given its solutions or the sum and product of roots

Doing the corresponding product

$$(x-\dfrac{1}{3})\cdot(x+\dfrac{2}{5})=x^2+\dfrac{1}{15}x-\dfrac{2}{15}$$ We can remove denominators by multiplying by $15$, so we have the equation $15x^2+x-2=0$.

If $D = 0$ means that the equation has a double root and, as this has to be $-6$, we will have $$(x-6)\cdot(x-6)=x^2-12x+36$$

Applying the formula $x^2-sx+p=0$, in this case $s = 9$ and $p = 20$.

$$x^2-9x+20=0$$ $$\displaystyle x=\frac{9 \pm \sqrt{81-80}}{2}= \frac{9 \pm 1}{2}=\left \{\begin{matrix} x_1=5 \\ x_2=4\end{matrix}\right.$$ So, the numbers we are looking for are $5$ and $4$.

Let's call $a$ and $b$ the sides of the rectangle. We know that $a\cdot b = 600$ and $2a + 2b = 100$, or, the same, $a + b = 50$.

Applying the formula $x^2-sx+p=0$ we will find that the corresponding quadratic equation is: $x^2-50x+600=0$. $$\displaystyle x=\frac{50 \pm \sqrt{50^2-4\cdot600}}{2}= \frac{50 \pm \sqrt{100}}{2}=\frac{50 \pm 10}{2}=\left\{\begin{matrix} x_1=30 \\ x_2=20\end{matrix}\right.$$ Then, the playground will be $30$ meters long and $20$ wide.