-16t^2+48t+15=0

Solution for -16t^2+48t+15=0 equation:

-16t^2+48t+15=0

a = -16; b = 48; c = +15;
Δ = b2-4ac
Δ = 482-4·(-16)·15
Δ = 3264
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:

$\sqrt{\Delta}=\sqrt{3264}=\sqrt{64*51}=\sqrt{64}*\sqrt{51}=8\sqrt{51}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-8\sqrt{51}}{2*-16}=\frac{-48-8\sqrt{51}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+8\sqrt{51}}{2*-16}=\frac{-48+8\sqrt{51}}{-32}
$