Question 613929
{{{  h(t) = -9.8t^2 -10t + 170 }}}
I'll plot this because the plot shows exactly what the
stone is doing at any time after it's thrown.
{{{ graph( 400, 400, -2, 5, -20, 250, -9.8x^2 - 10x + 170 ) }}}
You can see that at {{{ t = 0 }}}, the height, {{{ h(0) }}}, is 170.
What you are being asked is " How many seconds have passed
after stone is thrown until the stone hits the ground?"
Note that the stone is not thrown straight up and landing at
the 170 m height of the building. It is landing 170 m below this
at the ground level.
At ground level, {{{ h(t) = 0 }}}, so I can now say
{{{  -9.8t^2 - 10t + 170 = 0 }}}
Now I can use the quadratic formula to solve for {{{ t }}}
{{{t = (-b +- sqrt( b^2 - 4*a*c ))/(2*a) }}} 
{{{ a = -9.8 }}}
{{{ b = -10 }}}
{{{ c = 170 }}}
{{{ t = (-(-10) +- sqrt( (-10)^2-4*(-9.8)*170 ))/(2*(-9.8)) }}} 
{{{ t = ( 10 +- sqrt( 100 + 6664 ))/( -19.6 ) }}}
{{{ t = ( 10 +- sqrt( 6764 ))/( -19.6 ) }}} 
{{{ t = ( 10 - 82.244)/( -19.6 ) }}} 
{{{ t = (-72.244) / (-19.6) }}} 
{{{ t = 3.69 }}} sec ( rounded off )
The stone hits the ground in 3.69 seconds
Note that I had to use the negative root in the formula
because that gave me (-)/(-) or positive time.