32t^2-48=0

Solution for 32t^2-48=0 equation:

32t^2-48=0

a = 32; b = 0; c = -48;
Δ = b2-4ac
Δ = 02-4·32·(-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{-(0)-32\sqrt{6}}{2*32}=\frac{0-32\sqrt{6}}{64}
=-\frac{32\sqrt{6}}{64}
=-\frac{\sqrt{6}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{6}}{2*32}=\frac{0+32\sqrt{6}}{64}
=\frac{32\sqrt{6}}{64}
=\frac{\sqrt{6}}{2}
$