t^2-19t-20=0

Solution for t^2-19t-20=0 equation:

t^2-19t-20=0

a = 1; b = -19; c = -20;
Δ = b2-4ac
Δ = -192-4·1·(-20)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-21}{2*1}=\frac{-2}{2}
=-1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+21}{2*1}=\frac{40}{2}
=20
$