q^2+14q+13=0

Solution for q^2+14q+13=0 equation:

q^2+14q+13=0

a = 1; b = 14; c = +13;
Δ = b2-4ac
Δ = 142-4·1·13
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-12}{2*1}=\frac{-26}{2}
=-13
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+12}{2*1}=\frac{-2}{2}
=-1
$