15t^2-18t+2=0

Solution for 15t^2-18t+2=0 equation:

15t^2-18t+2=0

a = 15; b = -18; c = +2;
Δ = b2-4ac
Δ = -182-4·15·2
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{-(-18)-2\sqrt{51}}{2*15}=\frac{18-2\sqrt{51}}{30}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{51}}{2*15}=\frac{18+2\sqrt{51}}{30}
$