0=-3t^2+12t+5

Solution for 0=-3t^2+12t+5 equation:

0=-3t^2+12t+5

We move all terms to the left:

0-(-3t^2+12t+5)=0

We add all the numbers together, and all the variables

-(-3t^2+12t+5)=0

We get rid of parentheses

3t^2-12t-5=0

a = 3; b = -12; c = -5;
Δ = b2-4ac
Δ = -122-4·3·(-5)
Δ = 204
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:

$\sqrt{\Delta}=\sqrt{204}=\sqrt{4*51}=\sqrt{4}*\sqrt{51}=2\sqrt{51}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{51}}{2*3}=\frac{12-2\sqrt{51}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{51}}{2*3}=\frac{12+2\sqrt{51}}{6}
$