-2t^2+16=0

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

-2t^2+16=0

a = -2; b = 0; c = +16;
Δ = b2-4ac
Δ = 02-4·(-2)·16
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{2}}{2*-2}=\frac{0-8\sqrt{2}}{-4}
=-\frac{8\sqrt{2}}{-4}
=-\frac{2\sqrt{2}}{-1}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{2}}{2*-2}=\frac{0+8\sqrt{2}}{-4}
=\frac{8\sqrt{2}}{-4}
=\frac{2\sqrt{2}}{-1}
$