-t2+18=16

Solution for -t2+18=16 equation:

-t2+18=16

We move all terms to the left:

-t2+18-(16)=0

We add all the numbers together, and all the variables

-1t^2+2=0

a = -1; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-1)·2
Δ = 8
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:

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