T=-16t^2+105

Solution for T=-16t^2+105 equation:

=-16T^2+105

We move all terms to the left:

-(-16T^2+105)=0

We get rid of parentheses

16T^2-105=0

a = 16; b = 0; c = -105;
Δ = b2-4ac
Δ = 02-4·16·(-105)
Δ = 6720
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:

$\sqrt{\Delta}=\sqrt{6720}=\sqrt{64*105}=\sqrt{64}*\sqrt{105}=8\sqrt{105}$
$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{105}}{2*16}=\frac{0-8\sqrt{105}}{32}
=-\frac{8\sqrt{105}}{32}
=-\frac{\sqrt{105}}{4}
$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{105}}{2*16}=\frac{0+8\sqrt{105}}{32}
=\frac{8\sqrt{105}}{32}
=\frac{\sqrt{105}}{4}
$