-16t^2+575t+30=0

Solution for -16t^2+575t+30=0 equation:

-16t^2+575t+30=0

a = -16; b = 575; c = +30;
Δ = b2-4ac
Δ = 5752-4·(-16)·30
Δ = 332545
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(575)-\sqrt{332545}}{2*-16}=\frac{-575-\sqrt{332545}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(575)+\sqrt{332545}}{2*-16}=\frac{-575+\sqrt{332545}}{-32}
$