16t^2=6912

Solution for 16t^2=6912 equation:

16t^2=6912

We move all terms to the left:

16t^2-(6912)=0

a = 16; b = 0; c = -6912;
Δ = b2-4ac
Δ = 02-4·16·(-6912)
Δ = 442368
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:

$\sqrt{\Delta}=\sqrt{442368}=\sqrt{147456*3}=\sqrt{147456}*\sqrt{3}=384\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-384\sqrt{3}}{2*16}=\frac{0-384\sqrt{3}}{32}
=-\frac{384\sqrt{3}}{32}
=-12\sqrt{3}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+384\sqrt{3}}{2*16}=\frac{0+384\sqrt{3}}{32}
=\frac{384\sqrt{3}}{32}
=12\sqrt{3}
$