25t^2-60t+18=0

Solution for 25t^2-60t+18=0 equation:

25t^2-60t+18=0

a = 25; b = -60; c = +18;
Δ = b2-4ac
Δ = -602-4·25·18
Δ = 1800
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:

$\sqrt{\Delta}=\sqrt{1800}=\sqrt{900*2}=\sqrt{900}*\sqrt{2}=30\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-60)-30\sqrt{2}}{2*25}=\frac{60-30\sqrt{2}}{50}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-60)+30\sqrt{2}}{2*25}=\frac{60+30\sqrt{2}}{50}
$