3t^2-18t+5=0

Solution for 3t^2-18t+5=0 equation:

3t^2-18t+5=0

a = 3; b = -18; c = +5;
Δ = b2-4ac
Δ = -182-4·3·5
Δ = 264
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:

$\sqrt{\Delta}=\sqrt{264}=\sqrt{4*66}=\sqrt{4}*\sqrt{66}=2\sqrt{66}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{66}}{2*3}=\frac{18-2\sqrt{66}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{66}}{2*3}=\frac{18+2\sqrt{66}}{6}
$