1296-t^2=0

Solution for 1296-t^2=0 equation:

1296-t^2=0

We add all the numbers together, and all the variables

-1t^2+1296=0

a = -1; b = 0; c = +1296;
Δ = b2-4ac
Δ = 02-4·(-1)·1296
Δ = 5184
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5184}=72$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-72}{2*-1}=\frac{-72}{-2}
=+36
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+72}{2*-1}=\frac{72}{-2}
=-36
$