t2+20t+19=0

Solution for t2+20t+19=0 equation:

t2+20t+19=0

We add all the numbers together, and all the variables

t^2+20t+19=0

a = 1; b = 20; c = +19;
Δ = b2-4ac
Δ = 202-4·1·19
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-18}{2*1}=\frac{-38}{2}
=-19
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+18}{2*1}=\frac{-2}{2}
=-1
$