0=t^2-4-12

Solution for 0=t^2-4-12 equation:

0=t^2-4-12

We move all terms to the left:

0-(t^2-4-12)=0

We add all the numbers together, and all the variables

-(t^2-4-12)=0

We get rid of parentheses

-t^2+4+12=0

We add all the numbers together, and all the variables

-1t^2+16=0

a = -1; b = 0; c = +16;
Δ = b2-4ac
Δ = 02-4·(-1)·16
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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