g^2+15g+14=0

Solution for g^2+15g+14=0 equation:

g^2+15g+14=0

a = 1; b = 15; c = +14;
Δ = b2-4ac
Δ = 152-4·1·14
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-13}{2*1}=\frac{-28}{2}
=-14
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+13}{2*1}=\frac{-2}{2}
=-1
$