-16t2+42t+5.5=0

Solution for -16t2+42t+5.5=0 equation:

-16t^2+42t+5.5=0

a = -16; b = 42; c = +5.5;
Δ = b2-4ac
Δ = 422-4·(-16)·5.5
Δ = 2116
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2116}=46$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-46}{2*-16}=\frac{-88}{-32}
=2+3/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+46}{2*-16}=\frac{4}{-32}
=-1/8
$