9t^2-18t+1=0

Solution for 9t^2-18t+1=0 equation:

9t^2-18t+1=0

a = 9; b = -18; c = +1;
Δ = b2-4ac
Δ = -182-4·9·1
Δ = 288
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:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-12\sqrt{2}}{2*9}=\frac{18-12\sqrt{2}}{18}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+12\sqrt{2}}{2*9}=\frac{18+12\sqrt{2}}{18}
$