3+t^2+13t+13=0

Solution for 3+t^2+13t+13=0 equation:

3+t^2+13t+13=0

We add all the numbers together, and all the variables

t^2+13t+16=0

a = 1; b = 13; c = +16;
Δ = b2-4ac
Δ = 132-4·1·16
Δ = 105
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-\sqrt{105}}{2*1}=\frac{-13-\sqrt{105}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{105}}{2*1}=\frac{-13+\sqrt{105}}{2}
$