t^2-20t-5=0

Solution for t^2-20t-5=0 equation:

t^2-20t-5=0

a = 1; b = -20; c = -5;
Δ = b2-4ac
Δ = -202-4·1·(-5)
Δ = 420
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:

$\sqrt{\Delta}=\sqrt{420}=\sqrt{4*105}=\sqrt{4}*\sqrt{105}=2\sqrt{105}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{105}}{2*1}=\frac{20-2\sqrt{105}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{105}}{2*1}=\frac{20+2\sqrt{105}}{2}
$