16t^2-96+48=0

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

16t^2-96+48=0

We add all the numbers together, and all the variables

16t^2-48=0

a = 16; b = 0; c = -48;
Δ = b2-4ac
Δ = 02-4·16·(-48)
Δ = 3072
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:

$\sqrt{\Delta}=\sqrt{3072}=\sqrt{1024*3}=\sqrt{1024}*\sqrt{3}=32\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32\sqrt{3}}{2*16}=\frac{0-32\sqrt{3}}{32}
=-\frac{32\sqrt{3}}{32}
=-\sqrt{3}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{3}}{2*16}=\frac{0+32\sqrt{3}}{32}
=\frac{32\sqrt{3}}{32}
=\sqrt{3}
$