20t-5t^2-1=0

Solution for 20t-5t^2-1=0 equation:

20t-5t^2-1=0

a = -5; b = 20; c = -1;
Δ = b2-4ac
Δ = 202-4·(-5)·(-1)
Δ = 380
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:

$\sqrt{\Delta}=\sqrt{380}=\sqrt{4*95}=\sqrt{4}*\sqrt{95}=2\sqrt{95}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{95}}{2*-5}=\frac{-20-2\sqrt{95}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{95}}{2*-5}=\frac{-20+2\sqrt{95}}{-10}
$