-16t^2+10t+1453=0

Solution for -16t^2+10t+1453=0 equation:

-16t^2+10t+1453=0

a = -16; b = 10; c = +1453;
Δ = b2-4ac
Δ = 102-4·(-16)·1453
Δ = 93092
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:

$\sqrt{\Delta}=\sqrt{93092}=\sqrt{5476*17}=\sqrt{5476}*\sqrt{17}=74\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-74\sqrt{17}}{2*-16}=\frac{-10-74\sqrt{17}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+74\sqrt{17}}{2*-16}=\frac{-10+74\sqrt{17}}{-32}
$