a^2+13a+16=0

Solution for a^2+13a+16=0 equation:

a^2+13a+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:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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