-1=(6t^2)-18t+12

Solution for -1=(6t^2)-18t+12 equation:

-1=(6t^2)-18t+12

We move all terms to the left:

-1-((6t^2)-18t+12)=0

We get rid of parentheses

-6t^2+18t-12-1=0

We add all the numbers together, and all the variables

-6t^2+18t-13=0

a = -6; b = 18; c = -13;
Δ = b2-4ac
Δ = 182-4·(-6)·(-13)
Δ = 12
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:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{3}}{2*-6}=\frac{-18-2\sqrt{3}}{-12}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{3}}{2*-6}=\frac{-18+2\sqrt{3}}{-12}
$