g^2-16g+24=0

Solution for g^2-16g+24=0 equation:

g^2-16g+24=0

a = 1; b = -16; c = +24;
Δ = b2-4ac
Δ = -162-4·1·24
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{10}}{2*1}=\frac{16-4\sqrt{10}}{2}
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{10}}{2*1}=\frac{16+4\sqrt{10}}{2}
$