97=195t-16t^2

Solution for 97=195t-16t^2 equation:

97=195t-16t^2

We move all terms to the left:

97-(195t-16t^2)=0

We get rid of parentheses

16t^2-195t+97=0

a = 16; b = -195; c = +97;
Δ = b2-4ac
Δ = -1952-4·16·97
Δ = 31817
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-195)-\sqrt{31817}}{2*16}=\frac{195-\sqrt{31817}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-195)+\sqrt{31817}}{2*16}=\frac{195+\sqrt{31817}}{32}
$