16t^2-48t-64=0

Solution for 16t^2-48t-64=0 equation:

16t^2-48t-64=0

a = 16; b = -48; c = -64;
Δ = b2-4ac
Δ = -482-4·16·(-64)
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-80}{2*16}=\frac{-32}{32}
=-1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+80}{2*16}=\frac{128}{32}
=4
$