A question from Steward's Precalculus textbook 5th, Pg 55,
the original formula is $$h=\frac{1}{2}gt^2+V_0t$$
the question asks to write the formula in terms of $t$, the answer is $$t=\dfrac{-V_0\pm\sqrt{v_0^2+2gh}}{g}$$
I don't know the steps on how to get there
$$h=\frac{1}{2}gt^2+V_0t$$ $$2h=gt^2+2V_0t$$ $$gt^2+2V_0t-2h=0$$
now use this formula for Quadratic equation:$ax^2+bx+c=0$; $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$
so: $$t=\dfrac{-2V_0\pm\sqrt{4V_0^2+8gh}}{2g} $$ $$t=\dfrac{-V_0\pm\sqrt{V_0^2+2gh}}{g} $$