-16t^2-30t+2048=0

Solution for -16t^2-30t+2048=0 equation:

-16t^2-30t+2048=0

a = -16; b = -30; c = +2048;
Δ = b2-4ac
Δ = -302-4·(-16)·2048
Δ = 131972
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:

$\sqrt{\Delta}=\sqrt{131972}=\sqrt{4*32993}=\sqrt{4}*\sqrt{32993}=2\sqrt{32993}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-2\sqrt{32993}}{2*-16}=\frac{30-2\sqrt{32993}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+2\sqrt{32993}}{2*-16}=\frac{30+2\sqrt{32993}}{-32}
$