10=6t^2-18t+12

Solution for 10=6t^2-18t+12 equation:

10=6t^2-18t+12

We move all terms to the left:

10-(6t^2-18t+12)=0

We get rid of parentheses

-6t^2+18t-12+10=0

We add all the numbers together, and all the variables

-6t^2+18t-2=0

a = -6; b = 18; c = -2;
Δ = b2-4ac
Δ = 182-4·(-6)·(-2)
Δ = 276
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:

$\sqrt{\Delta}=\sqrt{276}=\sqrt{4*69}=\sqrt{4}*\sqrt{69}=2\sqrt{69}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{69}}{2*-6}=\frac{-18-2\sqrt{69}}{-12}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{69}}{2*-6}=\frac{-18+2\sqrt{69}}{-12}
$