g^2-16g+14=0

Solution for g^2-16g+14=0 equation:

g^2-16g+14=0

a = 1; b = -16; c = +14;
Δ = b2-4ac
Δ = -162-4·1·14
Δ = 200
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-10\sqrt{2}}{2*1}=\frac{16-10\sqrt{2}}{2}
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+10\sqrt{2}}{2*1}=\frac{16+10\sqrt{2}}{2}
$