2t^2-432=0

Solution for 2t^2-432=0 equation:

2t^2-432=0

a = 2; b = 0; c = -432;
Δ = b2-4ac
Δ = 02-4·2·(-432)
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*2}=\frac{0-24\sqrt{6}}{4}
=-\frac{24\sqrt{6}}{4}
=-6\sqrt{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24\sqrt{6}}{2*2}=\frac{0+24\sqrt{6}}{4}
=\frac{24\sqrt{6}}{4}
=6\sqrt{6}
$