-16t^2+263t+10=0

Solution for -16t^2+263t+10=0 equation:

-16t^2+263t+10=0

a = -16; b = 263; c = +10;
Δ = b2-4ac
Δ = 2632-4·(-16)·10
Δ = 69809
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{-(263)-\sqrt{69809}}{2*-16}=\frac{-263-\sqrt{69809}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(263)+\sqrt{69809}}{2*-16}=\frac{-263+\sqrt{69809}}{-32}
$