-16t^2+30t+350=0

Solution for -16t^2+30t+350=0 equation:

-16t^2+30t+350=0

a = -16; b = 30; c = +350;
Δ = b2-4ac
Δ = 302-4·(-16)·350
Δ = 23300
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:

$\sqrt{\Delta}=\sqrt{23300}=\sqrt{100*233}=\sqrt{100}*\sqrt{233}=10\sqrt{233}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-10\sqrt{233}}{2*-16}=\frac{-30-10\sqrt{233}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+10\sqrt{233}}{2*-16}=\frac{-30+10\sqrt{233}}{-32}
$