32t^2=27

Solution for 32t^2=27 equation:

32t^2=27

We move all terms to the left:

32t^2-(27)=0

a = 32; b = 0; c = -27;
Δ = b2-4ac
Δ = 02-4·32·(-27)
Δ = 3456
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:

$\sqrt{\Delta}=\sqrt{3456}=\sqrt{576*6}=\sqrt{576}*\sqrt{6}=24\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24\sqrt{6}}{2*32}=\frac{0-24\sqrt{6}}{64}
=-\frac{24\sqrt{6}}{64}
=-\frac{3\sqrt{6}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24\sqrt{6}}{2*32}=\frac{0+24\sqrt{6}}{64}
=\frac{24\sqrt{6}}{64}
=\frac{3\sqrt{6}}{8}
$