Rom

\(6x + 10 = x \\ 5x + 10 = 0 \\ 5x = -10 \\ x = -\dfrac {10}{5} = -2\)

\(-16 t^2 + 240 = 0\\ 16t^2 = 240 \\ t^2 = 15 \\ t = \pm \sqrt{15}\)

\(-\dfrac 3 4 x^2 + 2 = 0\)

multiply both sides by \(-\dfrac 4 3\)

\(x^2 - \dfrac 8 3 = 0\)

and thus \(c=-\dfrac 8 3\)

well...

if you want to get fancy the best unit for expressing the depth of water is the "fathom".  It's 6 feet or 72 inches

\(253440 in = \dfrac {253440in}{72 in/fathom}=3520 fathoms\)

\(10^{-\frac 9 2} = \left(10^{-9}\right)^{\frac 1 2} =\left(\dfrac 1 {10^9}\right)^{\frac 1 2}= \dfrac 1 {\sqrt{10^9}}\)

\(E = h \nu = (2.09\times 10^{16})(6.63 \times 10^{-34})= 1.38567\times 10^{-17} \mbox{J}\)

\(\log(1.38567\times 10^{-17}) = -16.86\)

\(u=f(x)\)

\(u^2 - 4 e^x u - 1 = 0\)

\(\left(u - 2 e^x\right)^2 - 4e^{2x}-1=0\)

\(\left(u-2e^x\right)^2=1+4e^{2x}\)

\(u-2e^x = \pm \sqrt{1+4e^{2x}}\)

\(u=2e^x \pm \sqrt{1+4e^{2x}}\)

\(f(x) = 2e^x \pm \sqrt{1+4e^{2x}}\)

\(f(0)=2\pm \sqrt{1+4}=2-\sqrt{5} \Rightarrow f(x) = 2e^x - \sqrt{1+4e^{2x}}\)

after 1 year you'll have 100-8=92% of your space left.  after 2 yrs, 92% of that, etc. after k years you have

area(k) = 575 (0.92)^k sq ft.

area(11) = 575 (0.92)¹¹​ = 229.7915 sq ft.

either click 2nd and then sin or

click sin and hold it a moment and it will change to asin which is the same as sin^-1

$${\mathtt{8.1}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{12}}}{\mathtt{\,\times\,}}{\mathtt{5.3}}{\mathtt{\,\times\,}}{\mathtt{10}} = {\mathtt{429\,300\,000\,000\,000}}$$