10=-3t^2+12t

Solution for 10=-3t^2+12t equation:

10=-3t^2+12t

We move all terms to the left:

10-(-3t^2+12t)=0

We get rid of parentheses

3t^2-12t+10=0

a = 3; b = -12; c = +10;
Δ = b2-4ac
Δ = -122-4·3·10
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{6}}{2*3}=\frac{12-2\sqrt{6}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{6}}{2*3}=\frac{12+2\sqrt{6}}{6}
$