-16t^2+42t+82=0

Solution for -16t^2+42t+82=0 equation:

-16t^2+42t+82=0

a = -16; b = 42; c = +82;
Δ = b2-4ac
Δ = 422-4·(-16)·82
Δ = 7012
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:

$\sqrt{\Delta}=\sqrt{7012}=\sqrt{4*1753}=\sqrt{4}*\sqrt{1753}=2\sqrt{1753}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-2\sqrt{1753}}{2*-16}=\frac{-42-2\sqrt{1753}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+2\sqrt{1753}}{2*-16}=\frac{-42+2\sqrt{1753}}{-32}
$