116(t)=-16t2+101

Solution for 116(t)=-16t2+101 equation:

116(t)=-16t^2+101

We move all terms to the left:

116(t)-(-16t^2+101)=0

We get rid of parentheses

16t^2+116t-101=0

a = 16; b = 116; c = -101;
Δ = b2-4ac
Δ = 1162-4·16·(-101)
Δ = 19920
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{19920}=\sqrt{16*1245}=\sqrt{16}*\sqrt{1245}=4\sqrt{1245}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(116)-4\sqrt{1245}}{2*16}=\frac{-116-4\sqrt{1245}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(116)+4\sqrt{1245}}{2*16}=\frac{-116+4\sqrt{1245}}{32}
$