19=t^2+5t+5

Solution for 19=t^2+5t+5 equation:

19=t^2+5t+5

We move all terms to the left:

19-(t^2+5t+5)=0

We get rid of parentheses

-t^2-5t-5+19=0

We add all the numbers together, and all the variables

-1t^2-5t+14=0

a = -1; b = -5; c = +14;
Δ = b2-4ac
Δ = -52-4·(-1)·14
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-9}{2*-1}=\frac{-4}{-2}
=+2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+9}{2*-1}=\frac{14}{-2}
=-7
$