t^2-40t-600=0

Solution for t^2-40t-600=0 equation:

t^2-40t-600=0

a = 1; b = -40; c = -600;
Δ = b2-4ac
Δ = -402-4·1·(-600)
Δ = 4000
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:

$\sqrt{\Delta}=\sqrt{4000}=\sqrt{400*10}=\sqrt{400}*\sqrt{10}=20\sqrt{10}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-20\sqrt{10}}{2*1}=\frac{40-20\sqrt{10}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+20\sqrt{10}}{2*1}=\frac{40+20\sqrt{10}}{2}
$