b^2+12b+5=0

Solution for b^2+12b+5=0 equation:

b^2+12b+5=0

a = 1; b = 12; c = +5;
Δ = b2-4ac
Δ = 122-4·1·5
Δ = 124
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{124}=\sqrt{4*31}=\sqrt{4}*\sqrt{31}=2\sqrt{31}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{31}}{2*1}=\frac{-12-2\sqrt{31}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{31}}{2*1}=\frac{-12+2\sqrt{31}}{2}
$