2t^2-18t+20=0

Solution for 2t^2-18t+20=0 equation:

2t^2-18t+20=0

a = 2; b = -18; c = +20;
Δ = b2-4ac
Δ = -182-4·2·20
Δ = 164
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:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{41}}{2*2}=\frac{18-2\sqrt{41}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{41}}{2*2}=\frac{18+2\sqrt{41}}{4}
$