-16t^2+558t=0

Solution for -16t^2+558t=0 equation:

-16t^2+558t=0

a = -16; b = 558; c = 0;
Δ = b2-4ac
Δ = 5582-4·(-16)·0
Δ = 311364
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{311364}=558$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(558)-558}{2*-16}=\frac{-1116}{-32}
=34+7/8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(558)+558}{2*-16}=\frac{0}{-32}
=0
$