-t^2-t+12=t0

Solution for -t^2-t+12=t0 equation:

-t^2-t+12=t0

We move all terms to the left:

-t^2-t+12-(t0)=0

We add all the numbers together, and all the variables

-1t^2-2t+12=0

a = -1; b = -2; c = +12;
Δ = b2-4ac
Δ = -22-4·(-1)·12
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{13}}{2*-1}=\frac{2-2\sqrt{13}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{13}}{2*-1}=\frac{2+2\sqrt{13}}{-2}
$