-16t^2+16t+370=0

Solution for -16t^2+16t+370=0 equation:

-16t^2+16t+370=0

a = -16; b = 16; c = +370;
Δ = b2-4ac
Δ = 162-4·(-16)·370
Δ = 23936
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:

$\sqrt{\Delta}=\sqrt{23936}=\sqrt{64*374}=\sqrt{64}*\sqrt{374}=8\sqrt{374}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{374}}{2*-16}=\frac{-16-8\sqrt{374}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{374}}{2*-16}=\frac{-16+8\sqrt{374}}{-32}
$