q^2+14q+4=0

Solution for q^2+14q+4=0 equation:

q^2+14q+4=0

a = 1; b = 14; c = +4;
Δ = b2-4ac
Δ = 142-4·1·4
Δ = 180
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-6\sqrt{5}}{2*1}=\frac{-14-6\sqrt{5}}{2}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+6\sqrt{5}}{2*1}=\frac{-14+6\sqrt{5}}{2}
$