-16t^2+43t+250=0

Solution for -16t^2+43t+250=0 equation:

-16t^2+43t+250=0

a = -16; b = 43; c = +250;
Δ = b2-4ac
Δ = 432-4·(-16)·250
Δ = 17849
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{-(43)-\sqrt{17849}}{2*-16}=\frac{-43-\sqrt{17849}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(43)+\sqrt{17849}}{2*-16}=\frac{-43+\sqrt{17849}}{-32}
$