-16t^2+144t+100=0

Solution for -16t^2+144t+100=0 equation:

-16t^2+144t+100=0

a = -16; b = 144; c = +100;
Δ = b2-4ac
Δ = 1442-4·(-16)·100
Δ = 27136
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:

$\sqrt{\Delta}=\sqrt{27136}=\sqrt{256*106}=\sqrt{256}*\sqrt{106}=16\sqrt{106}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(144)-16\sqrt{106}}{2*-16}=\frac{-144-16\sqrt{106}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(144)+16\sqrt{106}}{2*-16}=\frac{-144+16\sqrt{106}}{-32}
$