a^2+14a+5=0

Solution for a^2+14a+5=0 equation:

a^2+14a+5=0

a = 1; b = 14; c = +5;
Δ = b2-4ac
Δ = 142-4·1·5
Δ = 176
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{176}=\sqrt{16*11}=\sqrt{16}*\sqrt{11}=4\sqrt{11}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-4\sqrt{11}}{2*1}=\frac{-14-4\sqrt{11}}{2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+4\sqrt{11}}{2*1}=\frac{-14+4\sqrt{11}}{2}
$