t^2-18t+5=0

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

t^2-18t+5=0

a = 1; b = -18; c = +5;
Δ = b2-4ac
Δ = -182-4·1·5
Δ = 304
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:

$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-4\sqrt{19}}{2*1}=\frac{18-4\sqrt{19}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+4\sqrt{19}}{2*1}=\frac{18+4\sqrt{19}}{2}
$