-16t^2+88t+48=0

Solution for -16t^2+88t+48=0 equation:

-16t^2+88t+48=0

a = -16; b = 88; c = +48;
Δ = b2-4ac
Δ = 882-4·(-16)·48
Δ = 10816
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{10816}=104$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(88)-104}{2*-16}=\frac{-192}{-32}
=+6
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(88)+104}{2*-16}=\frac{16}{-32}
=-1/2
$