2h^2-16h+28=0

Solution for 2h^2-16h+28=0 equation:

2h^2-16h+28=0

a = 2; b = -16; c = +28;
Δ = b2-4ac
Δ = -162-4·2·28
Δ = 32
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{2}}{2*2}=\frac{16-4\sqrt{2}}{4}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{2}}{2*2}=\frac{16+4\sqrt{2}}{4}
$