0=-t^2+3t+12

Solution for 0=-t^2+3t+12 equation:

0=-t^2+3t+12

We move all terms to the left:

0-(-t^2+3t+12)=0

We add all the numbers together, and all the variables

-(-t^2+3t+12)=0

We get rid of parentheses

t^2-3t-12=0

a = 1; b = -3; c = -12;
Δ = b2-4ac
Δ = -32-4·1·(-12)
Δ = 57
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{57}}{2*1}=\frac{3-\sqrt{57}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{57}}{2*1}=\frac{3+\sqrt{57}}{2}
$