Originally Posted by

**chabmgph** Here is one way using the quadratic formula:

$\displaystyle 70=\frac{1.5x(20-x)}{2}$

...

$\displaystyle 140=1.5x(20-x)$

$\displaystyle 140=30x-1.5x^2$

$\displaystyle 1.5x^2-30x+140=0$ (In the form: $\displaystyle ax^2+bx+c=0$)

Then $\displaystyle a=1.5, b=-30, c=140$, Then using the quadratic formula:

$\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$,

we have

$\displaystyle x=\frac{-(-30)\pm \sqrt{(-30)^2-4(1.5)(140)}}{2(1.5)}$.

Thus $\displaystyle x=\frac{30\pm \sqrt{60}}{3}$

So $\displaystyle x \approx 12.58$ or $\displaystyle x \approx 7.42$