t^2-60t+600=0

Solution for t^2-60t+600=0 equation:

t^2-60t+600=0

a = 1; b = -60; c = +600;
Δ = b2-4ac
Δ = -602-4·1·600
Δ = 1200
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:

$\sqrt{\Delta}=\sqrt{1200}=\sqrt{400*3}=\sqrt{400}*\sqrt{3}=20\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-60)-20\sqrt{3}}{2*1}=\frac{60-20\sqrt{3}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-60)+20\sqrt{3}}{2*1}=\frac{60+20\sqrt{3}}{2}
$