t^2-20t-50=0

Solution for t^2-20t-50=0 equation:

t^2-20t-50=0

a = 1; b = -20; c = -50;
Δ = b2-4ac
Δ = -202-4·1·(-50)
Δ = 600
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:

$\sqrt{\Delta}=\sqrt{600}=\sqrt{100*6}=\sqrt{100}*\sqrt{6}=10\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-10\sqrt{6}}{2*1}=\frac{20-10\sqrt{6}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+10\sqrt{6}}{2*1}=\frac{20+10\sqrt{6}}{2}
$