t^2+40t+200=0

Solution for t^2+40t+200=0 equation:

t^2+40t+200=0

a = 1; b = 40; c = +200;
Δ = b2-4ac
Δ = 402-4·1·200
Δ = 800
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:

$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-20\sqrt{2}}{2*1}=\frac{-40-20\sqrt{2}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+20\sqrt{2}}{2*1}=\frac{-40+20\sqrt{2}}{2}
$