16t^2+96t+48=0

Solution for 16t^2+96t+48=0 equation:

16t^2+96t+48=0

a = 16; b = 96; c = +48;
Δ = b2-4ac
Δ = 962-4·16·48
Δ = 6144
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:

$\sqrt{\Delta}=\sqrt{6144}=\sqrt{1024*6}=\sqrt{1024}*\sqrt{6}=32\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-32\sqrt{6}}{2*16}=\frac{-96-32\sqrt{6}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+32\sqrt{6}}{2*16}=\frac{-96+32\sqrt{6}}{32}
$